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#### Page No 927:

#### Question 1:

Fill in the blanks:

(i) The probability of an impossible event is ....... .

(ii) The probability of a sure event is ........ .

(iii) For any event *E,* *P(E)* + *P* (not *E*) = ........ .

(iv) The probability of a possible but not a sure event lies between ......... and ........ .

(v) The sum of probabilities of all the outcomes of an experiment is ....... .

#### Answer:

(i) 0

(ii) 1

(iii) 1

(iv) 0, 1

(v) 1

#### Page No 927:

#### Question 2:

A coin is tossed once. What is the probability of getting a tail?

#### Answer:

When a coin is tossed once, the possible outcomes are H and T.

Total number of possible outcomes = 2

Favourable outcome = 1

∴ Probability of getting a tail = *P *(*T*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{1}{2}$

#### Page No 927:

#### Question 3:

Two coins are tossed simultaneously. Find the probability of getting

(i) exactly 1 head

(ii) at most 1 head

(iii) at least 1 head

#### Answer:

When two coins are tossed simultaneously, all possible outcomes are HH, HT, TH and TT.

Total number of possible outcomes = 4

(i) Let E be the event of getting exactly one head.

Number of favourable outcomes = 2

*P*(getting exactly 1 head) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{2}{4}=\frac{1}{2}$

(ii) Let E be the event of getting at most one head.

Number of favourable outcomes = 3

*P*(getting at most 1 head) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{3}{4}$

(iii) Let E be the event of getting at least one head.

Number of favourable outcomes = 3

*P*(getting at least 1 head) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{3}{4}$

#### Page No 928:

#### Question 4:

A die is thrown once. Find the probability of getting

(i) an even number

(ii) a number less than 5

(iii) a number greater than 2

(iv) a number between 3 and 6

(v) a number other than 3

(vi) the number 5.

#### Answer:

In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.

Total number of possible outcomes = 6

(i)

Let E be the event of getting an even number.

Then, the favourable outcomes are 2, 4 and 6.

Number of favourable outcomes = 3

∴ Probability of getting an even number = *P *(*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{3}{6}=\frac{1}{2}$

(ii)

Let E be the event of getting a number less than 5.

Then, the favourable outcomes are 1, 2, 3, 4

Number of favourable outcomes = 4

∴ Probability of getting a number less than 5 = *P *(*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{4}{6}=\frac{2}{3}$

(iii)

Let E be the event of getting a number greater than 2.

Then, the favourable outcomes are 3, 4, 5 and 6.

Number of favourable outcomes = 4

∴ Probability of getting a number greater than 2 = *P *(*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{4}{6}=\frac{2}{3}$

(iv)

Let E be the event of getting a number between 3 and 6.

∴ Probability of getting a number between 3 and 6 =

*P*(

*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{2}{6}=\frac{1}{3}$

(v)

Let E be the event of getting a number other than 3.

∴ Probability of getting a number other than 3 =

*P*(

*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{5}{6}$

(vi)

Let E be the event of getting the number 5.

∴ Probability of getting the number 5 =

*P*(

*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{1}{6}$

#### Page No 928:

#### Question 5:

A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is a consonant.

#### Answer:

Let E be the event of getting a consonant.

Out of 26 letters of English alphabets, there are 21 consonants.

∴ P(getting a consonant) = P(E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}\mathrm{E}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{21}{26}$

Thus, the probability of getting a consonant is $\frac{21}{26}$.

#### Page No 928:

#### Question 6:

A child has die whose 6 faces show the letters given below.

A | B | C | A | A | B |

The die is thrown once. What is the probability of getting (i) A, (ii) B?

#### Answer:

When the die is thrown once, then any one of the six faces can show up.

∴ Total number of outcomes = 6

(i) There are 3 faces on the die showing the letter A.

Favourable number of outcomes = 3

∴ P(Getting the letter A) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{3}{6}=\frac{1}{2}$

(ii) There are 2 faces on the die showing the letter B.

Favourable number of outcomes = 2

∴ P(Getting the letter B) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{2}{6}=\frac{1}{3}$

#### Page No 928:

#### Question 7:

12 defective pens are accidentally mixed up the 132 good ones. It is not possible to just look at a pen and tell whether it is defective or not. One pen is taken out at random from the lot. Find the probability that the pen taken out is (i) a good one, (ii) defective

#### Answer:

Number of defective pens in the lot = 12

Number of good pens in the lot = 132

Total number of pens in the lot = 12 + 132 = 144

∴ Total number of outcomes = 144

(i) There are 132 good pens in the lot. Out of these pens, one good pen can be taken out in 132 ways.

Favourable number of outcomes = 132

∴ P(Pen taken out is good one) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{132}{144}=\frac{11}{12}$

(ii) There are 12 defective pens in the lot. Out of these pens, one defective pen can be taken out in 12 ways.

Favourable number of outcomes = 12

∴ P(Pen taken out is defective) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{12}{144}=\frac{1}{12}$

#### Page No 928:

#### Question 8:

In a lottery there are 10 prizes and 25 blanks. What is the probability of getting a prize?

#### Answer:

Total number of lottery tickets = 10 + 25 = 35

Number of prizes = 10

Let E be the event of getting a prize.

∴ P(getting a prize) = P(E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}\mathrm{E}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{10}{35}=\frac{2}{7}$

Thus, the probability of getting a prize is $\frac{2}{7}$.

#### Page No 928:

#### Question 9:

250 lottery tickets were sold and there are 5 prizes on these tickets. If Kunal has purchased one lottery ticket, what is the probability that he wins a prize?

#### Answer:

Total number of tickets = 250

Kunal wins a prize if he gets a ticket that assures a prize.

Number of tickets on which prizes are assured = 5

∴ *P *(Kunal wins a prize) = $\frac{5}{250}=\frac{1}{50}$

#### Page No 928:

#### Question 10:

If the probability of winning a game is 0.7, what is the probability of losing it?

#### Answer:

For any event E, P(E) + P(not E) = 1

Let probability of winning a game = P(E) = 0.7

∴ P(winning a game) + P(losing a game) = 1

⇒ P(losing a game) = 1 − 0.7

= 0.3

Thus, the probability of losing a game is 0.3.

#### Page No 928:

#### Question 11:

A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the numbers 1, 2, 3,..., 12 as shown in the figure. What is the probability that it will point to

(i) 6?

(ii) an even number

(iii) a prime number?

(iv) a number which is a multiple of 5?

(v) a number which is a factor of 8?

#### Answer:

The arrow can come to rest at any one of the numbers 1, 2, 3, 4,..., 12.

∴ Total number of outcomes = 12

(i) There is only one 6 in the figure. So, there is only one way when the arrow will come to rest pointing to the number 6.

Favourable number of outcomes = 1

∴ P(Arrow will point to 6) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{1}{12}$

(ii) There are six even numbers in the figure. These are 2, 4, 6, 8, 10 and 12. So, there are 6 ways when the arrow will come to rest pointing to an even number.

Favourable number of outcomes = 6

∴ P(Arrow will point to an even number) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{6}{12}=\frac{1}{2}$

(iii) There are five prime numbers in the figure. These are 2, 3, 5, 7 and 11. So, there are 5 ways when the arrow will come to rest pointing to a prime number.

Favourable number of outcomes = 5

∴ P(Arrow will point to a prime number) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{5}{12}$

(iv) There are two multiples of 5 in the figure. These are 5 and 10. So, there are 2 ways when the arrow will come to rest pointing to a number which is a multiple of 5.

Favourable number of outcomes = 2

∴ P(Arrow will point to a number which is a multiple of 5) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{2}{12}=\frac{1}{6}$

(v) There are four factors of 8 in the figure. These are 1, 2, 4 and 8. So, there are 4 ways when the arrow will come to rest pointing to a number which is a factor of 8.

Favourable number of outcomes = 4

∴ P(Arrow will point to a number which is a factor of 8) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{4}{12}=\frac{1}{3}$

#### Page No 928:

#### Question 12:

17 cards numbered 1, 2, 3, 4, .... ,17 are put in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the card drawn bears (i) an odd number (ii) a number divisible by 5.

#### Answer:

Total number of cards = 17

(i) Let E_{1} be the event of choosing an odd number.

These numbers are 1, 3, 5, 7, 9, 11, 13, 15 and 17.

∴ P(getting an odd number) = P(E_{1}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_{1}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{9}{17}$

Thus, the probability that the card drawn bears an odd number is $\frac{9}{17}$.

(i) Let E_{2} be the event of choosing a number divisible by 5.

These numbers are 5, 10 and 15.

∴ P(getting a number divisible by 5) = P(E_{2}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_{2}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{3}{17}$

Thus, the probability that the card drawn bears a number divisible by 5 is $\frac{3}{17}$.

#### Page No 929:

#### Question 13:

A bag contains 4 white balls, 5 red balls, 2 black balls and 4 green balls. A ball is drawn at random from the bag. Find the probability that it is

(i) black,

(ii) not green,

(iii) red or white,

(iv) neither red nor green.

#### Answer:

Total number of balls = 15

(i) Number of black balls = 2

∴ P(getting a black ball) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{2}{15}$

Thus, the probability of getting a black ball is $\frac{2}{15}$.

(ii) Number of balls which are not green = 4 + 5 + 2 = 11

∴ P(getting a ball which is not green) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{11}{15}$

Thus, the probability of getting a ball which is not green is $\frac{11}{15}$.

(iii) Number of balls which are either red or white = 4 + 5 = 9

∴ P(getting a ball which is red or white) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{9}{15}=\frac{3}{5}$

Thus, the probability of getting a ball which is red or white is $\frac{3}{5}$.

(iv) Number of balls which are neither red nor green = 4 + 2 = 6

∴ P(getting a ball which is neither red nor green) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{6}{15}=\frac{2}{5}$

Thus, the probability of getting a ball which is neither red nor green is $\frac{2}{5}$.

#### Page No 929:

#### Question 14:

A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is (i) white or blue, (ii) neither white nor black.

#### Answer:

Number of white balls in the bag = 5

Number of red balls in the bag = 7

Number of black balls in the bag = 4

Number of blue balls in the bag = 2

Total number of balls in the bag = 5 + 7 + 4 + 2 = 18

∴ Total number of outcomes = 18

(i) There are 7 balls (5 white and 2 blue) in the bag which are either white or blue. So, there are 7 ways to draw a ball from the bag which is white or blue.

Favourable number of outcomes = 7

∴ P(Drawn ball is white or blue) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{7}{18}$

(ii) There are 9 balls (7 red and 2 blue) in the bag which are neither white nor black. So, there are 9 ways to draw a ball from the bag which is neither white nor black.

Favourable number of outcomes = 9

∴ P(Drawn ball is neither white nor black) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{9}{18}=\frac{1}{2}$

#### Page No 929:

#### Question 15:

A jar contains 24 marbles. Some of these are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is $\frac{2}{3}$. Find the number of blue marbles in the jar.

#### Answer:

Total number of marbles = 24.

Let the number of blue marbles be *x*.

Then, the number of green marbles = 24 − *x*

∴ P(getting a green marble) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{24-x}{24}$

But, P(getting a green marble) = $\frac{2}{3}$ (given)

$\therefore \frac{24-x}{24}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow 3(24-x)=48\phantom{\rule{0ex}{0ex}}\Rightarrow 72-3x=48\phantom{\rule{0ex}{0ex}}\Rightarrow 3x=72-48\phantom{\rule{0ex}{0ex}}\Rightarrow 3x=24\phantom{\rule{0ex}{0ex}}\Rightarrow x=8$

Thus, the number of blue marbles in the jar is 8.

#### Page No 929:

#### Question 16:

A jar contains 54 marbles, each of which some are blue, some are green and some are white. The probability of selecting a blue marble at random is $\frac{1}{3}$ and the probability of selecting a green marble at random is $\frac{4}{9}$. How many white marbles does the jar contain?

#### Answer:

Total number of marbles = 54.

It is given that, P(getting a blue marble) = $\frac{1}{3}$ and P(getting a green marble) = $\frac{4}{9}$

Let P(getting a white marble) be *x*.

Since, there are only 3 types of marbles in the jar, the sum of probabilities of all three marbles must be 1.

$\therefore \frac{1}{3}+\frac{4}{9}+x=1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3+4}{9}+x=1\phantom{\rule{0ex}{0ex}}\Rightarrow x=1-\frac{7}{9}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{2}{9}$

∴ P(getting a white marble) = $\frac{2}{9}$ ...(1)

Let the number of white marbles be *n*.

Then, P(getting a white marble) = $\frac{n}{54}$ ... (2)

From (1) and (2),

$\frac{n}{54}=\frac{2}{9}\phantom{\rule{0ex}{0ex}}\Rightarrow n=\frac{2\times 54}{9}\phantom{\rule{0ex}{0ex}}\Rightarrow n=12$

Thus, there are 12 white marbles in the jar.

#### Page No 929:

#### Question 17:

The probability of selecting a red ball at random from the jar that contains only red, blue and orange balls is $\frac{1}{4}$. The probability of selecting a blue ball at random from the same jar is $\frac{1}{3}$. If the jar contains 10 orange balls, find the total number of balls in the jar.

#### Answer:

It is given that,

P(getting a red ball) = $\frac{1}{4}$ and P(getting a blue ball) = $\frac{1}{3}$

Let P(getting an orange ball) be* x*.

Since, there are only 3 types of balls in the jar, the sum of probabilities of all the three balls must be 1.

$\therefore \frac{1}{4}+\frac{1}{3}+x=1\phantom{\rule{0ex}{0ex}}\Rightarrow x=1-\frac{1}{4}-\frac{1}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{12-3-4}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{5}{12}$\

∴ P(getting an orange ball) = $\frac{5}{12}$

Let the total number of balls in the jar be *n*.

∴ P(getting an orange ball) = $\frac{10}{n}$

$\Rightarrow \frac{10}{n}=\frac{5}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow n=24$

Thus, the total number of balls in the jar is 24.

#### Page No 929:

#### Question 18:

A bag contains 18 balls out of which *x* balls are red.

(i) If the ball is drawn at random from the bag, what is the probability that it is not red?

(ii) If two more red balls are put in the bag, the probability of drawing a red ball will be $\frac{9}{8}$ times the probability of drawing a red ball in the first case. Find the value of *x*.

#### Answer:

Total number of balls = 18.

Number of red balls = *x*.

(i) Number of balls which are not red = 18 − *x*

∴ P(getting a ball which is not red) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{18-x}{18}$

Thus, the probability of drawing a ball which is not red is $\frac{18-x}{18}$.

(ii) Now, total number of balls = 18 + 2 = 20.

Number of red balls now = *x* + 2.

P(getting a red ball now) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{x+2}{20}$

and P(getting a red ball in first case) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{x}{18}$

Since, it is given that probability of drawing a red ball now will be $\frac{9}{8}$ times the probability of drawing a red ball in the first case.

Thus, $\frac{x+2}{20}=\frac{9}{8}\times \frac{x}{18}$

$\Rightarrow 144(x+2)=180x\phantom{\rule{0ex}{0ex}}\Rightarrow 144x+288=180x\phantom{\rule{0ex}{0ex}}\Rightarrow 36x=288\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{288}{36}=8$

Thus, the value of *x* is 8.

#### Page No 929:

#### Question 19:

A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball find the number of black balls in the bag. [CBSE 2017]

#### Answer:

The number of white balls in a bag = 15.

Let the number of black balls in that bag be *x.*

Then, the total number of balls in bag = 15+*x*.

Now, P(black ball)=$\frac{x}{15+x}$and P(white ball)=$\frac{15}{15+x}$.

According to question, P(black ball) = 3 P(white ball)

$\Rightarrow \frac{x}{15+x}=3\times \frac{15}{15+x}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{15+x}=\frac{45}{15+x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=45.$

Hence, number of black balls in bag = *x* = 45.

#### Page No 929:

#### Question 20:

A box contains 90 discs, which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

(i) a two-digit number

(ii) a number divisible by 5

#### Answer:

Ans

#### Page No 929:

#### Question 21:

A bag contains some balls of which *x* are white, 2*x *are black are 3*x *are red. A ball is selected at random. What is the probability that it is (i) not red? (ii) white?

#### Answer:

Number of white balls in the bag = *x*

Number of black balls in the bag = 2*x*

Number of red balls in the bag = 3*x*

Total number of balls in the bag = *x* + 2*x* + 3*x* = 6*x*

∴ Total number of outcomes = 6*x*

(i) There are 3*x* non-red balls (*x* white balls and 2*x* black balls) in the bag. So, there are 3*x* ways to draw a ball from the bag which is not red.

Favourable number of outcomes = 3*x*

∴ P(Selected ball is not red) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{3x}{6x}=\frac{1}{2}$

(ii) There are *x* white balls in the bag. So, there are *x* ways to draw a ball from the bag which is white.

Favourable number of outcomes = *x*

∴ P(Selected ball is white) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{x}{6x}=\frac{1}{6}$

#### Page No 930:

#### Question 22:

Cards, marked with numbers 5 to 50, are placed in a box and mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is (i) a prime number less than 10 (ii) a perfect square.

#### Answer:

All possible outcomes are 5, 6, 7, 8...................50.

(i) Out of the given numbers, the prime numbers less than 10 are 5 and 7.

Let E

_{1}be the event of getting a prime number less than 10.

Then, number of favourable outcomes = 2

∴

*P*(getting a prime number less than 10) = $\frac{2}{46}=\frac{1}{23}$

(ii) Out of the given numbers, the perfect squares are 9, 16, 25, 36 and 49.

Let E

_{2}be the event of getting a perfect square.

Then, number of favourable outcomes = 5

∴

*P*(getting a perfect square) = $\frac{5}{46}$

#### Page No 930:

#### Question 23:

A box contains cords numbered from 1 to 20. A card is drawn at random from the box. Find the probability that the number on the drawn card is (i) a prime number, (ii) a composite number, (iii) a number divisible by 3.

#### Answer:

There are 20 cards in the box. One card can drawn at random from the box in 20 ways.

∴ Total number of outcomes = 20

(i) The prime number cards in the box are 2, 3, 5, 7, 11, 13, 17 and 19. There are 8 prime number cards in the box. So, there are 8 ways to draw a card from the box which is a prime number card.

Favourable number of outcomes = 8

∴ P(Number on the drawn card is a prime number) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{8}{20}=\frac{2}{5}$

(ii) The composite number cards in the box are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18 and 20. There are 11 composite number cards in the box. So, there are 11 ways to draw a card from the box which is a composite number card.

Favourable number of outcomes = 11

∴ P(Number on the drawn card is a composite number) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{11}{20}$

(iii) The cards with number divisible by 3 on them in the box are 3, 6, 9, 12, 15 and 18. There are 6 cards with number divisible by 3 on them in the box. So, there are 6 ways to draw a card from the box with number divisible by 3 on them.

Favourable number of outcomes = 6

∴ P(Number on the drawn card is a number divisible by 3) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{6}{20}=\frac{3}{10}$

#### Page No 930:

#### Question 24:

A box contains cards bearing numbers 6 to 70. If one card is drawn at random from the box, find the probability that it bears

(i) a one-digit number,

(ii) a number divisible by 5,

(iii) an odd number less than 30,

(iv) a composite number between 50 and 70.

#### Answer:

Given number 6, 7, 8, .... , 70 form an AP with *a* = 6 and *d* = 1.

Let *T _{n}* = 70. Then,

6 + (

*n*− 1)1 = 70

⇒ 6 +

*n*− 1 = 70

⇒

*n*= 65

Thus, total number of outcomes = 65.

(i) Let E

_{1}be the event of getting a one-digit number.

Out of these numbers, one-digit numbers are 6, 7, 8 and 9.

Number of favourable outcomes = 4.

∴ P(getting a one-digit number) = P(E

_{1}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_{1}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{4}{65}$

Thus, the probability that the card bears a one-digit number is $\frac{4}{65}$.

(ii) Let E

_{2}be the event of getting a number divisible by 5.

Out of these numbers, numbers divisible by 5 are 10, 15, 20, ... , 70.

Given number 10, 15, 20, .... , 70 form an AP with

*a*= 10 and

*d*= 5.

Let

*T*= 70. Then,

_{n}10 + (

*n*− 1)5 = 70

⇒ 10 + 5

*n*− 5 = 70

⇒ 5

*n*= 65

⇒

*n*= 13

Thus, number of favourable outcomes = 13.

∴ P(getting a number divisible by 5) = P(E

_{2}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_{2}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{13}{65}=\frac{1}{5}$

Thus, the probability that the card bears a number divisible by 5 is $\frac{1}{5}$.

(iii) Let E

_{3}be the event of getting an odd number less than 30.

Out of these numbers, odd numbers less than 30 are 7, 9, 11, ... , 29.

Given number 7, 9, 11, .... , 29 form an AP with

*a*= 7 and

*d*= 2.

Let

*T*= 29. Then,

_{n}7 + (

*n*− 1)2 = 29

⇒ 7 + 2

*n*− 2 = 29

⇒ 2

*n*= 24

⇒

*n*= 12

Thus, number of favourable outcomes = 12.

∴ P(getting an odd number less than 30) = P(E

_{3}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_{3}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{12}{65}$

Thus, the probability that the card bears an odd number less than 30 is $\frac{12}{65}$.

(iv) Let E

_{4}be the event of getting a composite number between 50 and 70.

Out of these numbers, composite numbers between 50 and 70 are 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68 and 69.

Number of favourable outcomes = 15.

∴ P(getting a composite number between 50 and 70) = P(E

_{4}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_{4}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{15}{65}=\frac{3}{13}$

Thus, the probability that the card bears a composite number between 50 and 70 is $\frac{3}{13}$.

#### Page No 930:

#### Question 25:

Cards marked with numbers 1, 3, 5, ..., 101 are placed in a bag and mixed thoroughly. A card is drawn at random from the bag. Find the probability that the number on the drawn card is

(i) less than 19,

(ii) a prime number less than 20.

#### Answer:

Given number 1, 3, 5, ..., 101 form an AP with *a* = 1 and *d* = 2.

Let *T _{n}* = 101. Then,

1 + (

*n*− 1)2 = 101

⇒ 1 + 2

*n*− 2 = 101

⇒ 2

*n*= 102

⇒

*n*= 51

Thus, total number of outcomes = 51.

(i) Let E

_{1}be the event of getting a number less than 19.

Out of these numbers, numbers less than 19 are 1, 3, 5, ... , 17.

Given number 1, 3, 5, .... , 17 form an AP with

*a*= 1 and

*d*= 2.

Let

*T*= 17. Then,

_{n}1 + (

*n*− 1)2 = 17

⇒ 1 + 2

*n*− 2 = 17

⇒ 2

*n*= 18

⇒

*n*= 9

Thus, number of favourable outcomes = 9.

∴ P(getting a number less than 19) = P(E

_{1}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_{1}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{9}{51}=\frac{3}{17}$

Thus, the probability that the number on the drawn card is less than 19 is $\frac{3}{17}$.

(ii) Let E

_{2}be the event of getting a prime number less than 20.

Out of these numbers, prime numbers less than 20 are 3, 5, 7, 11, 13, 17 and 19.

Thus, number of favourable outcomes = 7.

∴ P(getting a prime number less than 20) = P(E

_{2}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_{2}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{7}{51}$

Thus, the probability that the number on the drawn card is a prime number less than 20 is $\frac{7}{51}$.

#### Page No 930:

#### Question 26:

Cards bearing numbers 1, 3, 5, .... , 35 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing

(i) a prime number less than 15,

(ii) a number divisible by 3 and 5.

#### Answer:

Given number 1, 3, 5, .... , 35 form an AP with *a* = 1 and *d* = 2.

Let *T _{n}* = 35. Then,

1 + (

*n*− 1)2 = 35

⇒ 1 + 2

*n*− 2 = 35

⇒ 2

*n*= 36

⇒

*n*= 18

Thus, total number of outcomes = 18.

(i) Let E

_{1}be the event of getting a prime number less than 15.

Out of these numbers, prime numbers less than 15 are 3, 5, 7, 11 and 13.

Number of favourable outcomes = 5.

∴ P(getting a prime number less than 15) = P(E

_{1}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_{1}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{5}{18}$

Thus, the probability of getting a card bearing a prime number less than 15 is $\frac{5}{18}$.

(ii) Let E

_{2}be the event of getting a number divisible by 3 and 5.

Out of these numbers, number divisible by 3 and 5 means number divisible by 15 is 15.

Number of favourable outcomes = 1.

∴ P(getting a number divisible by 3 and 5) = P(E

_{2}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_{2}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{1}{18}$

Thus, the probability of getting a card bearing a number divisible by 3 and 5 is $\frac{1}{18}$.

#### Page No 930:

#### Question 27:

A box contains cards numbered 3, 5, 7, 9, .... , 35, 37. A card is drawn at random from the box. Find the probability that the number on the card is a prime number.

#### Answer:

Given numbers 3, 5, 7, 9, .... , 35, 37 form an AP with *a* = 3 and *d* = 2.

Let *T*_{n }= 37. Then,

3 + (*n* − 1)2 = 37

⇒ 3 + 2*n* − 2 = 37

⇒ 2*n* = 36

⇒ *n* = 18

Thus, total number of outcomes = 18.

Let E be the event of getting a prime number.

Out of these numbers, the prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 and 37.

Number of favourable outcomes = 11.

∴ P(getting a prime number) = P(E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}\mathrm{E}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{11}{18}$

Thus, the probability that the number on the card is a prime number is $\frac{11}{18}$.

#### Page No 930:

#### Question 28:

Cards numbered 1 to 30 are put in a bag. A card is drawn at random from the bag. Find the probability that the number on the drawn card is

(i) not divisible by 3,

(ii) a prime number greater than 7,

(iii) not a perfect square number.

#### Answer:

Total number of outcomes = 30.

(i) Let E_{1} be the event of getting a number not divisible by 3.

Out of these numbers, numbers divisible by 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27 and 30.

Number of favourable outcomes = 30 − 10 = 20

∴ P(getting a number not divisible by 3) = P(E_{1}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_{1}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{20}{30}=\frac{2}{3}$

Thus, the probability that the number on the card is not divisible by 3 is $\frac{2}{3}$.

(ii) Let E_{2} be the event of getting a prime number greater than 7.

Out of these numbers, prime numbers greater than 7 are 11, 13, 17, 19, 23 and 29.

Number of favourable outcomes = 6

∴ P(getting a prime number greater than 7) = P(E_{2}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_{2}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{6}{30}=\frac{1}{5}$

Thus, the probability that the number on the card is a prime number greater than 7 is $\frac{1}{5}$.

(iii) Let E_{3} be the event of getting a number which is not a perfect square number.

Out of these numbers, perfect square numbers are 1, 4, 9, 16 and 25.

Number of favourable outcomes = 30 − 5 = 25

∴ P(getting non-perfect square number) = P(E_{3}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_{3}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{25}{30}=\frac{5}{6}$

Thus, the probability that the number on the card is not a perfect square number is $\frac{5}{6}$.

#### Page No 930:

#### Question 29:

A box contains 25 cards numbered from 1 to 25. A card is drawn at random from the bag. Find the probability that the number on the drawn card is

(i) divisible by 2 or 3,

(ii) a prime number.

#### Answer:

Total number of outcomes = 25

(i) Let E_{1} be the event of getting a card divisible by 2 or 3.

Out of the given numbers, numbers divisible by 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22 and 24.

Out of the given numbers, numbers divisible by 3 are 3, 6, 9, 12, 15, 18, 21 and 24.

Out of the given numbers, numbers divisible by both 2 and 3 are 6, 12, 18 and 24.

Number of favourable outcomes = 16

∴ P(getting a card divisible by 2 or 3) = P(E_{1}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_{1}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{16}{25}$

Thus, the probability that the number on the drawn card is divisible by 2 or 3 is $\frac{16}{25}$.

(ii) Let E_{2} be the event of getting a prime number.

Out of the given numbers, prime numbers are 2, 3, 5, 7, 11, 13, 17, 19 and 23.

Number of favourable outcomes = 9

∴ P(getting a prime number) = P(E_{2}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}{\mathrm{E}}_{2}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{9}{25}$

Thus, the probability that the number on the drawn card is a prime number is $\frac{9}{25}$.

#### Page No 930:

#### Question 30:

Tickets numbered 2, 3, 4, 5, ..., 100, 101 are placed in a box and mixed thoroughly. One ticket is drawn at random from the box. Find the probability that the number on the ticket is

(i) an even number

(ii) a number less than 16

(iii) a number which is a perfect square

(iv) a prime number less than 40

#### Answer:

All possible outcomes are 2, 3, 4, 5................101.

Number of all possible outcomes = 100

(i) Out of these the numbers that are even = 2, 4, 6, 8...................100

Let E_{1} be the event of getting an even number.

Then, number of favourable outcomes = 50 $\left[{T}_{n}=100\Rightarrow 2+(n-1)\times 2=100,\Rightarrow n=50\right]\phantom{\rule{0ex}{0ex}}$

∴* P* (getting an even number) = $\frac{50}{100}=\frac{1}{2}$

(ii) Out of these, the numbers that are less than 16 = 2,3,4,5,..................15.

Let E_{2} be the event of getting a number less than 16.

Then, number of favourable outcomes = 14

∴* P* (getting a number less than 16) = $\frac{14}{100}=\frac{7}{50}$

(iii) Out of these, the numbers that are perfect squares = 4, 9,16,25, 36, 49, 64, 81 and 100

Let E_{3} be the event of getting a number that is a perfect square.

∴

*P*(getting a number that is a perfect square) = $\frac{9}{100}$

(iv) Out of these, prime numbers less than 40 = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37

Let E

_{4}be the event of getting a prime number less than 40.

Then, number of favourable outcomes = 12

∴

*P*(getting a prime number less than 40) = $\frac{12}{100}=\frac{3}{25}$

#### Page No 931:

#### Question 31:

In a family of 3 children, find the probability of having at least one boy.

#### Answer:

All possible outcomes are BBB, BBG, BGB, GBB, BGG, GBG, GGB and GGG.

Number of all possible outcomes = 8

Let E be the event of having at least one boy.

Then the outcomes are BBB, BBG, BGB, GBB, BGG, GBG and GGB.

Number of possible outcomes = 7

∴ P(Having at least one boy) = P(E) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{outcomes}\mathrm{favourable}\mathrm{to}\mathrm{E}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{7}{8}$

Thus, the probability of having at least one boy is $\frac{7}{8}$.

#### Page No 931:

#### Question 32:

Two different dice are thrown together. Find the probability that the numbers obtained have

(i) even sum

(ii) even product.

#### Answer:

Total number of possible outcomes is 36.

(i) The favorable outcomes are (1,1), (1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5), (4,2), (4,4), (4,6), (5,1), (5,3), (5,5), (6,2), (6,4), (6,6).

P(the sum is even)=$\frac{18}{36}=\frac{1}{2}.$

(ii) The favorable outcomes are (1,2), (1,4), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,2), (3,4), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,2), (5,4), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6).

P(the product is even)=$\frac{27}{36}=\frac{3}{4}.$

#### Page No 931:

#### Question 33:

Two different dice are thrown together. Find the probability that the numbers obtained

(i) have a sum less than 7

(ii) have a product less than 16

(iii) is a doublet of odd numbers. [CBSE 2017]

#### Answer:

Total number of possible outcomes is 36.

(i) The favorable outcomes for sum of numbers on dices less than 7 are (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (4,1), (4,2), (5,1).

P(the sum of numbers appeared is less than 7)=$\frac{15}{36}=\frac{5}{12}.$

(ii) The favorable outcomes for product of numbers on dices less than 16 are (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (6,1), (6,2).

P(the product of numbers appeared is less than 16)=$\frac{25}{36}.$

(iii) The favorable outcomes are (1,1), (3,3), (5,5).

P(the doublet of odd numbers)=$\frac{3}{36}=\frac{1}{12}.$

#### Page No 931:

#### Question 34:

Peter throws two different dice together and finds the product of the two numbers obtained. Rina throws a die and squares the number obtained. Who has the better chance to get the number 25? [CBSE 2017]

#### Answer:

Total number of possible outcomes for Peter = 36.

Possible outcomes for Peter to get product 25 is (5,5).

Total number of possible outcomes for Rina = 6.

Possible outcomes for Rina to get the number whose square is 25 is 5.

Now, P(Peter will get 25)=$\frac{1}{36}.$

And, P(Rina will get 25)=$\frac{1}{6}.$

Since $\frac{1}{36}<\frac{1}{6}$, So Rina has better chances of getting 25.

#### Page No 931:

#### Question 35:

Find the probability of getting the sum of two numbers, less than 3 or more than 11, when a pair of distinct dice is thrown together. [CBSE 2017]

#### Answer:

Favorable outcomes for sum of numbers on dice less than 3 or more than 11 are (1,1), (6,6).

Required probability = P(sum is less than 3 or more than 11) = $\frac{2}{36}=\frac{1}{18}.$

#### Page No 931:

#### Question 36:

The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18. What is the number of rotten apples in the heap? [CBSE 2017]

#### Answer:

Total number of apples = 900.

P(a rotten apple) = 0.18

$\Rightarrow \frac{\mathrm{number}\mathrm{of}\mathrm{rotten}\mathrm{apples}}{\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{apples}}=0.18\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{number}\mathrm{of}\mathrm{rotten}\mathrm{apples}}{900}=0.18\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{The}\mathrm{number}\mathrm{of}\mathrm{rotten}\mathrm{apples}=0.18\times 900=162\mathrm{apples}.$

#### Page No 931:

#### Question 37:

What is the probability that an ordinary year has 53 Mondays?

#### Answer:

An ordinary year has 365 days consisting of 52 weeks and 1 day.

This day can be any day of the week.

∴ P(of this day to be Monday) = $\frac{1}{7}$

Thus, the probability that an ordinary year has 53 Mondays is $\frac{1}{7}$.

#### Page No 931:

#### Question 38:

Find the probability that a leap year selected at random will contain 53 Sundays.

#### Answer:

A leap year has 366 days with 52 weeks and 2 days.

Now, 52 weeks conatins 52 sundays.

The remaining two days can be:

(i) Sunday and Monday

(ii) Monday and Tuesday

(iii) Tuesday and Wednesday

(iv) Wednesday and Thursday

(v) Thursday and Friday

(vi) Friday and Saturday

(vii) Saturday and Sunday

Out of these 7 cases, there are two cases favouring it to be Sunday.

∴ P(a leap year having 53 Sundays) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{2}{7}$

Thus, the probability that a leap year selected at random will contain 53 Sundays is $\frac{2}{7}$.

#### Page No 931:

#### Question 39:

A letter is chosen at random from the letters of the word ASSOCIATION. Find the probability that the chosen letter is a

(i) vowel

(ii) consonant

(iii) an S.

#### Answer:

Total numbers of letters in the given word ASSOCIATION = 11

(i) Number of vowels ( A, O, I, A, I, O) in the given word = 6

∴* P *(getting a vowel) = $\frac{6}{11}$

(ii) Number of consonants in the given word ( S, S, C, T, N) = 5

∴* P *(getting a consonant) = $\frac{5}{11}$

(iii) Number of S in the given word = 2

∴ *P* (getting an S) = $\frac{2}{11}$

#### Page No 931:

#### Question 40:

A game consists of tossing a one-rupee coin three times, and noting its outcome each time. If getting the same result in all the tosses is a success, find the probability of losing the game.

#### Answer:

When a coin is tossed three times, the possible outcomes are

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

Total number of outcomes = 8

A game can be lost if the if all the tosses do not give the same result. The outcomes where tosses do not give the same result are

HHT, HTH, THH, HTT, THT, TTH

Favourable number of outcomes for losing the game = 6

∴ P(Game is lost) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}\mathrm{for}\mathrm{losing}\mathrm{the}\mathrm{game}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{6}{8}=\frac{3}{4}$

Thus, the probability of losing the game is $\frac{3}{4}$.

#### Page No 931:

#### Question 41:

The king, the jack and the 10 of spades are lost from a pack of 52 cards and a card is drawn from the remaining cards after shuffling. Find the probability of getting a

(i) red card

(ii) black jack

(iii) red king

(iv) 10 of hearts. [CBSE 2017]

#### Answer:

Total number of cards in a deck is 52.

The number of cards left after loosing the King, the Jack and the 10 of spade = 52$-$3 = 49.

(i) Red card left after loosing King, the Jack and the 10 of spade will be 13 hearts + 13 diamond = 26 cards

P(red card)=$\frac{26}{49}.$

(ii) Black jack left after loosing King, the Jack and the 10 of spade will be one only of club.

P(black jack)=$\frac{1}{49}.$

(iii) Red king left after loosing King, the Jack and the 10 of spade will be red of diamond and red of hearts i.e two cards.

P(red king)=$\frac{2}{49}.$

(iv) 10 of hearts will be one only after loosing King, the Jack and the 10 of spade

P(10 of hearts)=$\frac{1}{49}.$

#### Page No 931:

#### Question 42:

All red face cards are removed from a pack of playing cards. The remaining cards are well shuffled and then a card is drawn at random from them. Find the probability that the drawn card is

(i) a red card,

(ii) a face card,

(iii) a card of clubs.

#### Answer:

There are 6 red face cards. These are removed.

Thus, remaining number of cards = 52 − 6 = 46.

(i) Number of red cards now = 26 − 6 = 20.

∴ P(getting a red card) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{20}{46}=\frac{10}{23}$

Thus, the probability that the drawn card is a red card is $\frac{10}{23}$.

(ii) Number of face cards now = 12 − 6 = 6.

∴ P(getting a face card) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{6}{46}=\frac{3}{23}$

Thus, the probability that the drawn card is a face card is $\frac{3}{23}$.

(iii) Number of card of clubs = 12.

∴ P(getting a card of clubs) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{12}{46}=\frac{6}{23}$

Thus, the probability that the drawn card is a card of clubs is $\frac{6}{23}$.

#### Page No 932:

#### Question 43:

All kings, queens and aces are removed from a pack of 52 cards. The remaining cards are well-shuffled and then a card is drawn from it. Find the probability that the drawn card is

(i) a black face card,

(ii) a red card.

#### Answer:

There are 4 kings, 4 queens and 4 aces. These are removed.

Thus, remaining number of cards = 52 − 4 − 4 − 4 = 40.

(i) Number of black face cards now = 2 (only black jacks).

∴ P(getting a black face card) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{2}{40}=\frac{1}{20}$

Thus, the probability that the drawn card is a black face card is $\frac{1}{20}$.

(ii) Number of red cards now = 26 − 6 = 20.

∴ P(getting a red card) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{20}{40}=\frac{1}{2}$

Thus, the probability that the drawn card is a red card is $\frac{1}{2}$.

#### Page No 932:

#### Question 44:

A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability that the card drawn is neither a red card nor a queen.

#### Answer:

Total number of all possible outcomes= 52

There are 26 red cards (including 2 queens) and apart from these, there are 2 more queens.

Number of cards, each one of which is either a red card or a queen = 26 + 2 = 28

Let *E* be the event that the card drawn is neither a red card nor a queen.

Then, the number of favourable outcomes = (52 − 28) = 24

∴ *P*( getting neither a red card nor a queen) = *P* (*E*) = $\frac{24}{52}=\frac{6}{13}$

#### Page No 932:

#### Question 45:

Five cards − the ten, jack, queen, king and ace of diamonds are well shuffled with their faces downwards. One card is then picked up at random.

(a) What is the probability that the drawn card is the queeen?

(b) If the queen is drawn and put aside and a second card is drawn, find the probability that the second card is (i) an ace, (ii) a queen.

#### Answer:

Total number of cards = 5.

(a) Number of queens = 1.

∴ P(getting a queen) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{1}{5}$

Thus, the probability that the drawn card is the queen is $\frac{1}{5}$.

(b) When the queen is put aside, number of remaining cards = 4.

(i) Number of aces = 1.

∴ P(getting an ace) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{1}{4}$

Thus, the probability that the drawn card is an ace is $\frac{1}{4}$.

(ii) Number of queens = 0.

∴ P(getting a queen now) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{0}{4}=0$

Thus, the probability that the drawn card is a queen is 0.

#### Page No 932:

#### Question 46:

A card is drawn at random form a well-shuffled deck of playing cards. Find the probability that the card drawn is

(i) a card of spades of an ace

(ii) a red king

(iii) either a king or a queen

(iv) neither a king nor a queen.

#### Answer:

Total number of all possible outcomes= 52

(i) Number of spade cards = 13

Number of aces = 4 (including 1 of spade)

*P*( getting a spade or an ace card) = $\frac{16}{52}=\frac{4}{13}$

(ii) Number of red kings = 2

∴

*P*( getting a red king) = $\frac{2}{52}=\frac{1}{26}$

(iii) Total number of kings = 4

Total number of queens = 4

Let

*E*be the event of getting either a king or a queen.

Then, the favourable outcomes = 4 + 4 = 8

∴

*P*( getting a king or a queen) =

*P*(

*E*) = $\frac{8}{52}=\frac{2}{13}$

(iv) Let

*E*be the event of getting either a king or a queen. Then, ( not

*E*) is the event that drawn card is neither a king nor a

*P*(getting a king or a queen ) = $\frac{2}{13}$

Now,

*P*(

*E*) +

*P*(not

*E*) = 1

∴

*P*(getting neither a king nor a queen ) = $1-\left(\frac{2}{13}\right)=\frac{11}{13}$

#### Page No 932:

#### Question 47:

Find the probability that a number selected at random from the numbers 3, 4, 4,4 5, 5, 6, 6, 6, 7 will be their mean.

#### Answer:

The number are 3, 4, 4, 4, 5, 5, 6, 6, 6, 7. One number can be selected at random from the given 10 numbers in 10 ways.

Total number of outcomes = 10

Mean of the given numbers = $\frac{3+4+4+4+5+5+6+6+6+7}{10}=\frac{50}{10}=5$

Now, there are two 5's among the given numbers. So, there are 2 ways to select the number 5 i.e. the mean of the given numbers.

Favourable number of outcomes = 2

∴ P(Selected number is the mean of the given number) = P(Selecting the number 5) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{2}{10}=\frac{1}{5}$

Thus, the probability that a number selected at random from the given numbers will be their mean is $\frac{1}{5}$.

#### Page No 944:

#### Question 1:

If P(E) denotes the probability of an event E then [CBSE 2013C]

(a) P(E) < 0

(b) P(E) > 1

(c) 0 ≤ P(E) ≤ 1

(d) −1 ≤ P(E) ≤ 1

#### Answer:

Since, number of elements in the set of favourable cases is less than or equal to the number of elements in the set of whole number of cases,

their ratio always end up being 1 or less than 1.

Also, their ratio can never be negative.

Thus, probability of an event always lies between 0 and 1.

i.e. 0 ≤ P(E) ≤ 1

Hence, the correct answer is option (c).

#### Page No 944:

#### Question 2:

If the probability of occurence of an event is *p* then the probability of non-happening of this event is [CBSE 2013C]

(a) (*p* − 1)

(b) (1 − *p*)

(c) *p*

(d) $\left(1-\frac{1}{p}\right)$

#### Answer:

P(occurence of an event) = *p*

P(non-occurence of this event) = 1 − *p*

Hence, the correct answer is option (b).

#### Page No 944:

#### Question 3:

What is the probability of an impossible event?

(a) $\frac{1}{2}$

(b) 0

(c) 1

(d) none of these

#### Answer:

(b) 0

The probability of an impossible event is 0.

#### Page No 944:

#### Question 4:

What is the probability of a sure event?

(a) 0

(b) $\frac{1}{2}$

(c) 1

(d) none of these

#### Answer:

(c) 1

The probability of a sure event is 1.

#### Page No 944:

#### Question 5:

Which of the following cannot be the probability of an event? [CBSE 2013C]

(a) 1.5

(b) $\frac{3}{5}$

(c) 25%

(d) 0.3

#### Answer:

The probability of an event cannot be greater than 1.

Thus, the probability of an event cannot be 1.5.

Hence, the correct answer is option (a).

#### Page No 944:

#### Question 6:

A number is selected at random from the numbers 1 to 30. What is the probability that the selected number is a prime number? [CBSE 2014]

(a) $\frac{2}{3}$

(b) $\frac{1}{6}$

(c) $\frac{1}{3}$

(d) $\frac{11}{30}$

#### Answer:

Total number of outcomes = 30.

Prime numbers in 1 to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.

Number of favourable outcomes = 10.

∴ P(getting a prime number) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{10}{30}=\frac{1}{3}$

Thus, the probability that the selected number is a prime number is $\frac{1}{3}$.

Hence, the correct answer is option (c).

#### Page No 944:

#### Question 7:

The probability that a number selected at random from the numbers 1, 2, 3, ..., 15 is a multiple of 4, is [CBSE 2014]

(a) $\frac{4}{15}$

(b) $\frac{2}{15}$

(c) $\frac{1}{5}$

(d) $\frac{1}{3}$

#### Answer:

Total number of outcomes = 15.

Out of the given numbers, multiples of 4 are 4, 8 and 12.

Numbers of favourable outcomes = 3.

∴ P(getting a multiple of 4) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{3}{15}=\frac{1}{5}$

Thus, the probability that a number selected is a multiple of 4 is $\frac{1}{5}$.

Hence, the correct answer is option (c).

#### Page No 944:

#### Question 8:

A box conatins cards numbered 6 to 50. A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect square is [CBSE 2013]

(a) $\frac{1}{45}$

(b) $\frac{2}{15}$

(c) $\frac{4}{45}$

(d) $\frac{1}{9}$

#### Answer:

Total number of outcomes = 45.

Out of the given numbers, perfect square numbers are 9, 16, 25, 36 and 49.

Numbers of favourable outcomes = 5.

∴ P(getting a number which is a perfect square) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{5}{45}=\frac{1}{9}$

Thus, the probability that the drawn card has a number which is a perfect square is $\frac{1}{9}$.

Hence, the correct answer is option (d).

#### Page No 945:

#### Question 9:

A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears prime number less than 23 is [CBSE 2013]

(a) $\frac{7}{90}$

(b) $\frac{1}{9}$

(c) $\frac{4}{15}$

(d) $\frac{8}{89}$

#### Answer:

Total number of discs = 90.

Out of the given numbers, prime numbers less than 23 are 2, 3, 5, 7, 11, 13, 17 and 19.

Numbers of favourable outcomes = 8.

∴ P(getting a prime number which is less than 23) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{8}{90}=\frac{4}{45}$

Thus, the probability that the disc bears prime number less than 23 is $\frac{4}{45}$.

**Disclaimer**: There is a misprinting in the question. If they ask for the probability of all prime numbers less than 90, then we get $\frac{4}{15}$.

Hence, the correct answer is option (c).

#### Page No 945:

#### Question 10:

Cards bearing numbers 2, 3, 4, ..., 11 are kept in a bag. A card is drawn at random from the bag. The probability of getting a card with a prime number is [CBSE 2012]

(a) $\frac{1}{2}$

(b) $\frac{2}{5}$

(c) $\frac{3}{10}$

(d) $\frac{5}{9}$

#### Answer:

Total number of cards = 10.

Out of the given numbers, prime numbers are 2, 3, 5, 7 and 11.

Numbers of favourable outcomes = 5.

∴ P(getting a prime number) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{5}{10}=\frac{1}{2}$

Thus, the probability of getting a card with a prime number is $\frac{1}{2}$.

Hence, the correct answer is option (a).

#### Page No 945:

#### Question 11:

One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number, which is a multiple of 7, is [CBSE 2013C]

(a) $\frac{1}{7}$

(b) $\frac{1}{8}$

(c) $\frac{1}{5}$

(d) $\frac{7}{40}$

#### Answer:

Total number of tickets = 40.

Out of the given numbers, multiples of 7 are 7, 14, 21, 28 and 35.

Numbers of favourable outcomes = 5.

∴ P(getting a number which is a multiple of 7) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{5}{40}=\frac{1}{8}$

Thus, the probability that the selected ticket has a number, which is a multiple of 7, is $\frac{1}{8}$.

Hence, the correct answer is option (b).

#### Page No 945:

#### Question 12:

Which of the following cannot be the probability of an event?

(a) $\frac{1}{3}$

(b) 0.3

(c) 33%

(d) $\frac{7}{6}$

#### Answer:

(d) $\frac{7}{6}$

Explanation:

Probability of an event can't be more than 1.

#### Page No 945:

#### Question 13:

If the probability of winning a game is 0.4, the probability of losing it is

(a) 0.96

(b) $\frac{1}{0.4}$

(c) 0.6

(d) none of these

#### Answer:

(c) 0.6

Explanation:

Let *E* be the event of winning a game.

Then, ( not *E*) is the event of not winning the game or of losing the game.

Then, *P*(*E*) = 0.4

Now, *P*(*E*) + *P*(not *E*) = 1

⇒ 0.4 + *P*(not *E*) = 1

⇒ *P*(not *E*) = 1− 0.4 = 0.6

∴ *P*(losing the game) = *P*(not *E*) = 0.6

#### Page No 945:

#### Question 14:

If an event cannot occur then its probability is

(a) 1

(b) $\frac{1}{2}$

(c) $\frac{3}{4}$

(d) 0

#### Answer:

(d) 0

If an event cannot occur, its probability is 0.

#### Page No 945:

#### Question 15:

There are 20 tickets numbered as 1, 2, 3,..., 20 respectively. One ticket is drawn at random. What is the probability that the number on the ticket drawn is a multiple of 5?

(a) $\frac{1}{4}$

(b) $\frac{1}{5}$

(c) $\frac{2}{5}$

(d) $\frac{3}{10}$

#### Answer:

(b) $\frac{1}{5}$

Explanation:

Total number of tickets = 20

Out of the given ticket numbers, multiples of 5 are 5, 10, 15 and 20.

Number of favourable outcomes = 4

∴ *P*(getting a multiple of 5) = $\frac{4}{20}=\frac{1}{5}$

#### Page No 945:

#### Question 16:

There are 25 tickets numbered 1, 2, 3, 4,..., 25 respectively. One ticket is draw at random. What is the probability that the number on the ticket is a multiple of 3 or 5?

(a) $\frac{2}{5}$

(b) $\frac{11}{25}$

(c) $\frac{12}{25}$

(d) $\frac{13}{25}$

#### Answer:

(c) $\frac{12}{25}$

Explanation:

Total number of tickets = 25

Let E be the event of getting a multiple of 3 or 5.

Then,

Multiples of 3 are 3, 6, 9, 12, 15, 18, 21 and 24.

Multiples of 5 are 5, 10, 15, 20 and 25.

Number of favourable outcomes = ( 8 + 5 − 1) = 12

∴ *P *(getting a multiple of 3 or 5 ) = *P* (E) = $\frac{12}{25}$

#### Page No 945:

#### Question 17:

Cards, each marked with one of the numbers 6, 7, 8,..., 15, are placed in a box and mixed thoroughly. One card is drawn at random from the box. What is the probability of getting a card with a number less than 10?

(a) $\frac{3}{5}$

(b) $\frac{1}{3}$

(c) $\frac{1}{2}$

(d) $\frac{2}{5}$

#### Answer:

(d) $\frac{2}{5}$

Explanation:

All possible outcomes are 6, 7, 8................15.

Number of all possible outcomes = 10

Number of favourable outcomes = 4

∴

*P*(getting a number that is less than 10) = $\frac{4}{10}=\frac{2}{5}$

#### Page No 946:

#### Question 18:

A die is thrown once. The probability of getting an even number is [CBSE 2013]

(a) $\frac{1}{2}$

(b) $\frac{1}{3}$

(c) $\frac{1}{6}$

(d) $\frac{5}{6}$

#### Answer:

Total number of outcomes = 6.

Out of the given numbers, even numbers are 2, 4 and 6.

Numbers of favourable outcomes = 3.

∴ P(getting an even number) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{3}{6}=\frac{1}{2}$

Thus, the probability of getting an even number is $\frac{1}{2}$.

Hence, the correct answer is option (a).

#### Page No 946:

#### Question 19:

The probability of throwing a number greater than 2 with a fair die is [CBSE 2011]

(a) $\frac{2}{5}$

(b) $\frac{5}{6}$

(c) $\frac{1}{3}$

(d) $\frac{2}{3}$

#### Answer:

Total number of outcomes = 6.

Out of the given numbers, numbers greater than 2 are 3, 4, 5 and 6.

Numbers of favourable outcomes = 4.

∴ P(getting a number greater than 2) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{4}{6}=\frac{2}{3}$

Thus, the probability of getting a number greater than 2 is $\frac{2}{3}$.

Hence, the correct answer is option (d).

#### Page No 946:

#### Question 20:

A die is thrown once. The probability of getting an odd number greater than 3 is [CBSE 2013C]

(a) $\frac{1}{3}$

(b) $\frac{1}{6}$

(c) $\frac{1}{2}$

(d) 0

#### Answer:

Total number of outcomes = 6.

Out of the given numbers, odd number greater than 3 is 5.

Numbers of favourable outcomes = 1.

∴ P(getting an odd number greater than 3) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{1}{6}$

Thus, the probability of getting an odd number greater than 3 is $\frac{1}{6}$.

Hence, the correct answer is option (b).

#### Page No 946:

#### Question 21:

A die is thrown once. The probability of getting a prime number is

(a) $\frac{2}{3}$

(b) $\frac{1}{3}$

(c) $\frac{1}{2}$

(d) $\frac{1}{6}$

#### Answer:

(c) $\frac{1}{2}$

Explanation:

In a single throw of a die, the possible outcomes are:

1, 2, 3, 4, 5, 6

Total number of possible outcomes = 6

Let E be the event of getting a prime number.

Then, the favourable outcomes are 2, 3 and 5.

Number of favourable outcomes = 3

∴ Probability of getting a prime number = *P *(*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favorable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{3}{6}=\frac{1}{2}$

#### Page No 946:

#### Question 22:

Two dice are thrown together. The probability of getting the same number on the both dice is [CBSE 2012]

(a) $\frac{1}{2}$

(b) $\frac{1}{3}$

(c) $\frac{1}{6}$

(d) $\frac{1}{12}$

#### Answer:

Total number of outcomes = 36.

Getting the same number on both the dice means (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).

Numbers of favourable outcomes = 6.

∴ P(getting the same number on both dice) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{6}{36}=\frac{1}{6}$

Thus, the probability of getting the same number on both dice is $\frac{1}{6}$.

Hence, the correct answer is option (c).

#### Page No 946:

#### Question 23:

The probability of getting 2 heads, when two coins are tossed, is [CBSE 2012]

(a) 1

(b) $\frac{3}{4}$

(c) $\frac{1}{2}$

(d) $\frac{1}{4}$

#### Answer:

All possible outcomes are HH, HT, TH and TT.

Total number of outcomes = 4.

Getting 2 heads means getting HH.

Numbers of favourable outcomes = 1.

∴ P(getting 2 heads) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Number}\mathrm{of}\mathrm{all}\mathrm{possible}\mathrm{outcomes}}$

= $\frac{1}{4}$

Thus, the probability of getting 2 heads is $\frac{1}{4}$.

Hence, the correct answer is option (d).

#### Page No 946:

#### Question 24:

Two dice are thrown together. The probability of getting a doublet is

(a) $\frac{1}{3}$

(b) $\frac{1}{6}$

(c) $\frac{1}{4}$

(d) $\frac{2}{3}$

#### Answer:

(2) $\frac{1}{6}$

Explanation:

When two dice are thrown simultaneously, all possible outcomes are:

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

Number of all possible outcomes = 36

Let *E* be the event of getting a doublet.

Then the favourable outcomes are:

(1,1), (2,2) , (3,3) (4,4), (5,5), (6,6)

Number of favourable outcomes = 6

∴ *P*(getting a doublet) = *P* ( *E*) = $\frac{6}{36}=\frac{1}{6}$

#### Page No 946:

#### Question 25:

Two coins are tossed simultaneously. What is the probability of getting at most one head?

(a) $\frac{1}{4}$

(b) $\frac{1}{2}$

(c) $\frac{2}{3}$

(d) $\frac{3}{4}$

#### Answer:

(d) $\frac{3}{4}$

Explanation:

When two coins are tossed simultaneously, the possible outcomes are HH, HT, TH and TT.

Total number of possible outcomes = 4

Let E be the event of getting at most one head.

Number of favourable outcomes = 3

*P*(getting at most 1head) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{3}{4}$

#### Page No 946:

#### Question 26:

Three coins are tossed simultaneously. What is the probability of getting exactly two heads?

(i) $\frac{1}{2}$

(ii) $\frac{1}{4}$

(iii) $\frac{3}{8}$

(iv) $\frac{3}{4}$

#### Answer:

(c) $\frac{3}{8}$

Explanation:

When 3 coins are tossed simultaneously, the possible outcomes are HHH, HHT, HTH, THH, THT, HTT, TTH and TTT.

Total number of possible outcomes = 8

Let E be the event of getting exactly two heads.

Number of favourable outcomes = 3

*P*(

*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favo}\text{u}\mathrm{rable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}\mathrm{outcomes}}=\frac{3}{8}$

#### Page No 947:

#### Question 27:

In a lottery, there are 8 prizes and 16 blanks. What is the probability of getting a prize?

(a) $\frac{1}{2}$

(b) $\frac{1}{3}$

(c) $\frac{2}{3}$

(d) none of these

#### Answer:

(b) $\frac{1}{3}$

Explanation:

Number of prizes = 8

Number of blanks = 16

Total number of tickets = 8 +16= 24

∴ *P*(getting a prize ) = $\frac{8}{24}=\frac{1}{3}$

#### Page No 947:

#### Question 28:

In a lottery, there are 6 prizes and 24 blanks. What is the probability of not getting a prize?

(a) $\frac{3}{4}$

(b) $\frac{3}{5}$

(c) $\frac{4}{5}$

(d) none of these

#### Answer:

(c) $\frac{4}{5}$

Explanation:

Number of prizes = 6

Number of blanks = 24

Total number of tickets = 6 + 24= 30

∴ *P*(not getting a prize ) = $\frac{24}{30}=\frac{4}{5}$

#### Page No 947:

#### Question 29:

A box contains 3 blue, 2 white and 4 red marbles. If a marbles. If a marble is drawn at random from the box, what is the probability that it will not be a white marble?

(a) $\frac{1}{3}$

(b) $\frac{4}{9}$

(c) $\frac{7}{9}$

(d) $\frac{2}{9}$

#### Answer:

(c) $\frac{7}{9}$

Explanation:

Total possible outcomes = Total number of marbles

It means the marble can be either blue or red but not white.

Number of favourable outcomes = (3 + 4) = 7 marbles

*P*(not getting a white marble ) = $\frac{7}{9}$

#### Page No 947:

#### Question 30:

A bag contains 4 red and 6 black balls. A ball is taken out of the bag at random. What is the probability of getting a black ball?

(a) $\frac{2}{5}$

(b) $\frac{3}{5}$

(c) $\frac{1}{10}$

(d) none of these

#### Answer:

(b)$\frac{3}{5}$

Explanation:

Total possible outcomes = Total number of balls = ( 4 + 6) = 10

Number of black balls = 6

∴ *P *(getting a black ball ) = $\frac{6}{10}=\frac{3}{5}$

#### Page No 947:

#### Question 31:

A bag contains 8 red, 2 black and 5 white balls. One ball is drawn at random. What is the probability that the ball drawn is not black?

(a) $\frac{18}{15}$

(b) $\frac{2}{15}$

(c) $\frac{13}{15}$

(d) $\frac{1}{3}$

#### Answer:

(c)$\frac{13}{15}$

Explanation:

Total possible outcomes = Total number of balls

∴

*P*(getting a ball that is not black) = $\frac{13}{15}$

#### Page No 947:

#### Question 32:

A bag contains 3 white, 4 red and 5 black balls. One ball is drawn at random. What is the probability that the ball drawn is neither black nor white?

(a) $\frac{1}{4}$

(b) $\frac{1}{2}$

(c) $\frac{1}{3}$

(d) $\frac{3}{4}$

#### Answer:

(c) $\frac{1}{3}$

Explanation:

Total possible outcomes = Total number of balls

∴

*P*(getting a ball that is neither black nor white) = $\frac{4}{12}=\frac{1}{3}$

#### Page No 947:

#### Question 33:

A card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a black king?

(a) $\frac{1}{13}$

(b) $\frac{1}{26}$

(c) $\frac{2}{39}$

(d) none of these

#### Answer:

(b) $\frac{1}{26}$

Explanation:

Total number of all possible outcomes = 52

Number of black kings = 2

∴ *P*( getting a black king) = $\frac{2}{52}=\frac{1}{26}$

#### Page No 947:

#### Question 34:

From a well-shuffled deck of 52 cards, one card is drawn at random. What is the probability of getting a queen?

(a) $\frac{1}{13}$

(b) $\frac{1}{26}$

(c) $\frac{4}{39}$

(d) none of these

#### Answer:

(a) $\frac{1}{13}$

Explanation:

Total number of all possible outcomes= 52

Number of queens = 4

∴ *P*( getting a queen) = $\frac{4}{52}=\frac{1}{13}$

#### Page No 947:

#### Question 35:

One card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a face card?

(a) $\frac{1}{26}$

(b) $\frac{3}{26}$

(c) $\frac{3}{13}$

(d) $\frac{4}{13}$

#### Answer:

(c) $\frac{3}{13}$

Explanation:

Total number of all possible outcomes= 52

Number of face cards ( 4 kings + 4 queens + 4 jacks) = 12

∴ *P*( getting a face card) = $\frac{12}{52}=\frac{3}{13}$

#### Page No 948:

#### Question 36:

Once card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a black face card?

(a) $\frac{1}{26}$

(b) $\frac{3}{26}$

(c) $\frac{3}{13}$

(d) $\frac{3}{14}$

#### Answer:

(b) $\frac{3}{26}$

Explanation:

Total number of all possible outcomes= 52

Number of black face cards ( 2 kings + 2 queens + 2 jacks) = 6

∴ *P*( getting a black face card) = $\frac{6}{52}=\frac{3}{26}$

#### Page No 948:

#### Question 37:

One card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a 6?

(a) $\frac{3}{26}$

(b) $\frac{1}{52}$

(c) $\frac{1}{13}$

(d) none of these

#### Answer:

(c) $\frac{1}{13}$

Explanation:

Total number of all possible outcomes = 52

Number of 6 in a deck of 52 cards = 4

∴ *P*( getting a 6) = $\frac{4}{52}=\frac{1}{13}$

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