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Page No 157:
Question 1:
Choose the correct answer from the given four options:
In the formula $\overline{x}=a+\frac{{f}_{i}{d}_{i}}{{f}_{i}}$, for finding the mean of grouped data d_{i}’s are deviations from a of
(A) lower limits of the classes
(B) upper limits of the classes
(C) mid points of the classes
(D) frequencies of the class marks
Answer:
We know that, d_{i} = x_{i} $$ a, where, x_{i} are data and a is the assumed mean
i.e. d_{i} are the deviations from of mid$$points of the classes.
Hence, the correct answer is option (C).
Page No 157:
Question 2:
Choose the correct answer from the given four options:
While computing mean of grouped data, we assume that the frequencies are
(A) evenly distributed over all the classes
(B) centred at the classmarks of the classes
(C) centred at the upper limits of the classes
(D) centred at the lower limits of the classes
Answer:
In grouping the data all the observations between lower and upper limits of class marks are taken in one group then mid value or class mark is taken for further calculation.
In computing the mean of grouped data, the frequencies are centered at the class marks of the classes.
Hence, the correct answer is option (B).
Page No 157:
Question 3:
Choose the correct answer from the given four options:
If x_{i}’s are the mid points of the class intervals of grouped data, f_{i}’s are the corresponding frequencies and $\overline{)x}$ is the mean, then $\sum _{}$$\left({f}_{i}{x}_{i}\overline{)x}\right)$ is equal to
(A) 0
(B) –1
(C) 1
(D) 2
Answer:
We know that,
$\sum _{i=1}^{n}\left({f}_{i}\times {x}_{i}\right)$ = $n\times \overline{x}$ .....(1)
$\sum _{i=1}^{n}\overline{x}=\overline{x}+\overline{x}+\overline{x}+\overline{x}.....ntimes\phantom{\rule{0ex}{0ex}}$
$\Rightarrow \sum _{i=1}^{n}\overline{x}=n\overline{x}\phantom{\rule{0ex}{0ex}}$ .....(2)
From (1) and (2)
$\sum _{i=1}^{n}\left({f}_{i}\times {x}_{i}\right)$ = $\sum _{i=1}^{n}\overline{x}\phantom{\rule{0ex}{0ex}}$
$\Rightarrow $$\sum _{}$$\left({f}_{i}{x}_{i}\overline{)x}\right)$ = 0
Hence, the correct answer is option (A).
Page No 157:
Question 4:
Choose the correct answer from the given four options:
In the formula $\overline{)x}=a+h\frac{{f}_{i}{u}_{i}}{{f}_{i}}$, for finding the mean of grouped frequency distribution, u_{i} =
(A) $\frac{{x}_{i}+a}{h}$
(B) h (x_{i }– a)
(C) $\frac{{x}_{i}a}{h}$
(D) $\frac{a{x}_{i}}{h}$
Answer:
Above formula is a step deviation (shortcut) formula.
Where x_{i} are data values, a is assumed mean and h is class size, when class size is same we simplify the calculations of the mean by computing the coded mean of u_{1}, u_{2}, u_{3}…..where
${u}_{i}=\frac{{x}_{i}a}{h}$
Hence, the correct answer is option (C).
Page No 158:
Question 5:
Choose the correct answer from the given four options:
The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its
(A) mean
(B) median
(C) mode
(D) all the three above
Answer:
The intersection point of less than ogive and more than ogive gives the median on the abscissa.
Hence, the correct answer is option (B).
Page No 158:
Question 6:
Choose the correct answer from the given four options:
For the following distribution :
Class  0 – 5  5 – 10  10 – 15  15 – 20  20 – 25 
Frequency  10  15  12  20  9 
the sum of lower limits of the median class and modal class is
(A) 15
(B) 25
(C) 30
(D) 35
Answer:
Here
Class  Frequency  Cumulative Frequency 
05  10  10 
510  15  25 
1015  12  37 
1520  20  57 
2025  9  66 
Now, $\frac{N}{2}$ = $\frac{66}{2}$ = 33, which lies in the interval 10  15.
Therefore, lower limit of the median class is 10.
The highest frequency is 20, which lies in the interval 15  20.
Therefore, lower limit of modal class is 15.
Therefore, required sum is 10 + 15 = 25.
Hence, the correct answer is option (B).
Page No 158:
Question 7:
Choose the correct answer from the given four options:
Consider the following frequency distribution :
Class  0 – 5  6 – 11  12 – 17  18 – 23  24 – 29 
Frequency  13  10  15  8  11 
The upper limit of the median class is
(A) 17
(B) 17.5
(C) 18
(D) 18.5
Answer:
Given, classes are not continuous, so we make continuous by subtracting 0.5 from lower limit and adding 0.5 to upper limit of each class.
Class  Frequency  Cumulative Frequency 
0.5 – 5.5  13  13 
5.5 – 11.5  10  23 
11.5 – 17.5  15  38 
17.5 – 23.5  8  46 
23.5 – 28.5  11  57 
Here, $\frac{N}{2}$ = $\frac{57}{2}$ = 28.5 which lies in the interval 11.5 – 17.5.
$\therefore $ the upper limit is 17.5.
Hence, the correct answer is option (B).
Page No 158:
Question 8:
Choose the correct answer from the given four options:
For the following distribution :
Marks  Number of students 
Below 10  3 
Below 20  12 
Below 30  27 
Below 40  57 
Below 50  75 
Below 60

80

the modal class is
(A) 10 – 20
(B) 20 – 30
(C) 30 – 40
(D) 50 – 60
Answer:
Marks  Number of Students  Cumulative Frequency 
Below 10  3  3 
10 $$ 20  (12 – 3) = 9  12 
20 $$ 30  (27 – 12) = 15  27 
30 $$ 40  (57 – 27) = 30  57 
40 $$ 50  (75 – 57) = 18  75 
50 $$ 60  (80 – 75) = 5  80 
Here, we see that the highest frequency is 30, which lies in the interval 30 – 40.
Hence, the correct answer is option (C).
Page No 158:
Question 9:
Choose the correct answer from the given four options:
Consider the data :
Class  65 – 85  85 – 105  105 – 125  125 – 145  145 – 165  165 – 185  185 – 205 
Frequency  4  5  13  20  14  7  4 
The difference of the upper limit of the median class and the lower limit of the modal class is
(A) 0
(B) 19
(C) 20
(D) 38
Answer:
Class  Frequency  Cumulative Frequency 
65 – 85  4  4 
85 – 105  5  9 
105 – 125  13  22 
125 – 145  20  42 
145 – 165  14  56 
165 – 185  7  63 
185 – 205  4  67 
Here, $\frac{N}{2}$ = $\frac{67}{2}$ = 33.5 which lies in the interval 125 – 145.
Hence, upper limit of median class is 145.
Here, we see that the highest frequency is 20 which lies in 125 – 145. Hence, the lower limit of modal class is 125.
∴ Required difference = Upper limit of median class – Lower limit of modal class = 145 – 125 = 20
Hence, the correct answer is option (C).
Page No 159:
Question 10:
Choose the correct answer from the given four options:
The times, in seconds, taken by 150 atheletes to run a 110 m hurdle race are tabulated below :
Class  13.8 – 14  14 – 14.2  14.2 – 14.4  14.4 – 14.6  14.6 – 14.8  14.8 – 15 
Frequency  2  4  5  71  48  20 
The number of atheletes who completed the race in less then 14.6 seconds is :
(A) 11
(B) 71
(C) 82
(D) 130
Answer:
The number of athletes who completed the race in less than 14.6 = 2 + 4 + 5 + 71 = 82
Hence, the correct answer is option (C).
Page No 159:
Question 11:
Choose the correct answer from the given four options:
Consider the following distribution :
Marks obtained  Number of students 
More than or equal to 0  63 
More than or equal to 10  58 
More than or equal to 20  55 
More than or equal to 30  51 
More than or equal to 40  48 
More than or equal to 50  42 
(A) 3
(B) 4
(C) 48
(D) 51
Answer:
Marks obtained  Number of students  Frequency 
0 $$ 10  (63 $$ 58) = 5  5 
10 $$ 20  (58 $$ 55) = 3  3 
20 $$ 30  (55 $$ 51) = 4  4 
30 $$ 40  (51 $$ 48) = 3  3 
40 $$ 50  (48 $$ 42) = 6  6 
50...  42  42 
Frequency in the class interval 30 $$ 40 is 3.
Hence, the correct answer is option (A).
Page No 159:
Question 12:
Choose the correct answer from the given four options:
If an event cannot occur, then its probability is
(A) 1
(B)$\frac{3}{4}$
(C) $\frac{1}{2}$
(D) 0
Answer:
The event which cannot occur is said to be impossible event and probability of impossible event is zero.
Hence, the correct answer is option (D).
Page No 159:
Question 13:
Choose the correct answer from the given four options:
Which of the following cannot be the probability of an event?
(A) $\frac{1}{3}$
(B) 0.1
(C) 3%
(D) $\frac{17}{16}$
Answer:
Since, probability of an event always lies between 0 and 1, therefore probability of any event cannot be more than 1 i.e. it cannot be $\frac{17}{16}$.
Hence, the correct answer is option (D).
Page No 159:
Question 14:
Choose the correct answer from the given four options:
An event is very unlikely to happen. Its probability is closest to
(A) 0.0001
(B) 0.001
(C) 0.01
(D) 0.1
Answer:
The probability of an event which is very unlikely to happen is closest to zero and from the given options 0.0001 is closest to zero.
Hence, the correct answer is option (A).
Page No 159:
Question 15:
Choose the correct answer from the given four options:
If the probability of an event is p, the probability of its complementary event will be
(A) p – 1
(B) p
(C) 1 – p
(D) $1\frac{1}{p}$
Answer:
Since, probability of an event + probability of its complementary event = 1 so,
Probability of its complementary event = (1 – Probability of an event) = 1 – p
Hence, the correct answer is option (C).
Page No 160:
Question 16:
Choose the correct answer from the given four options:
The probability expressed as a percentage of a particular occurrence can never be
(A) less than 100
(B) less than 0
(C) greater than 1
(D) anything but a whole number
Answer:
We know that the probability expressed as a percentage always lies between 0 and 100. So, it cannot be less than 0.
Hence, the correct answer is option (B).
Page No 160:
Question 17:
Choose the correct answer from the given four options:
If P(A) denotes the probability of an event A, then
(A) P(A) < 0
(B) P(A) > 1
(C) 0 ≤ P(A) ≤ 1
(D) –1 ≤ P(A) ≤1
Answer:
Probability of an event always lies between 0 and 1.
Hence, the correct answer is option (C).
Page No 160:
Question 18:
Choose the correct answer from the given four options:
A card is selected from a deck of 52 cards. The probability of its being a red face card is
(A) $\frac{3}{26}$
(B) $\frac{3}{13}$
(C) $\frac{2}{13}$
(D) $\frac{1}{2}$
Answer:
In a deck of 52 cards, there are 12 face cards i.e. 6 red and 6 black cards.
So, probability of getting a red face card = $\frac{6}{52}$ = $\frac{3}{26}$.
Hence, the correct answer is option (A).
Page No 160:
Question 19:
Choose the correct answer from the given four options:
The probability that a non leap year selected at random will contain 53 sundays is
(A) $\frac{1}{7}$
(B) $\frac{2}{7}$
(C) $\frac{3}{7}$
(D) $\frac{5}{7}$
Answer:
A non  leap year has 365 days and therefore 52 weeks and 1 day. This 1 day may be Sunday or Monday or Tuesday or Wednesday or Thursday or Friday or Saturday.
Thus, out of 7 possibilities, 1 favorable event is the event that the one day is Sunday.
∴ Required probability = $\frac{1}{7}$
Hence, the correct answer is option (A).
Page No 160:
Question 20:
Choose the correct answer from the given four options:
When a die is thrown, the probability of getting an odd number less than 3 is
(A) $\frac{1}{6}$
(B) $\frac{1}{3}$
(C) $\frac{1}{2}$
(D) 0
Answer:
When a die is thrown, then total number of outcomes = 6
Odd number less than 3 is 1 only.
Number of possible outcomes = 1
∴ Required probability = $\frac{1}{6}$
Hence, the correct answer is option (A).
Page No 160:
Question 21:
Choose the correct answer from the given four options:
A card is drawn from a deck of 52 cards. The event E is that card is not an ace of hearts. The number of outcomes favourable to E is
(A) 4
(B) 13
(C) 48
(D) 51
Answer:
In a deck of 52 cards, there are 13 cards of heart and 1 is ace of heart.
Therefore, the number of outcomes favorable to E = 52 $$ 1 = 51
Hence, the correct answer is option (D).
Page No 160:
Question 22:
Choose the correct answer from the given four options:
The probability of getting a bad egg in a lot of 400 is 0.035. The number of bad eggs in the lot is
(A) 7
(B) 14
(C) 21
(D) 28
Answer:
Here, total number of eggs = 400
Probability of getting a bad egg = 0.035
$\Rightarrow \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{bad}\mathrm{egg}=\frac{\mathrm{Number}\mathrm{of}\mathrm{bad}\mathrm{eggs}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{eggs}}$
$\Rightarrow 0.035=\frac{\mathrm{Number}\mathrm{of}\mathrm{bad}\mathrm{eggs}}{400}$
∴ Number of bad eggs = 0.035 × 400 = 14
Hence, the correct answer is option (B).
Page No 160:
Question 23:
Choose the correct answer from the given four options:
A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many tickets has she bought?
(A) 40
(B) 240
(C) 480
(D) 750
Answer:
Given, total number of sold tickets = 6000
Let us assume that girl bought x tickets.
Then, probability of her winning the first prize is given as,
$\frac{x}{6000}=0.08$
∴ x = 480
Hence, the correct answer is option (C).
Page No 160:
Question 24:
Choose the correct answer from the given four options:
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 5 is
(A) $\frac{1}{5}$
(B) $\frac{3}{5}$
(C) $\frac{4}{5}$
(D) $\frac{1}{3}$
Answer:
Number of total outcomes = 40
Multiples of 5 between 1 to 40 = 5, 10, 15, 20, 25, 30, 35, 40
∴ Total number of possible outcomes = 8
∴ Required probability = $\frac{8}{40}$ = $\frac{1}{5}$
Hence, the correct answer is Option (A).
Page No 160:
Question 25:
Choose the correct answer from the given four options:
Someone is asked to take a number from 1 to 100. The probability that it is a prime is
(A) $\frac{1}{5}$
(B) $\frac{6}{25}$
(C) $\frac{1}{4}$
(D) $\frac{13}{50}$
Answer:
Total numbers of outcomes = 100
So, the prime numbers between 1 to 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 56, 61, 67, 71, 73, 79, 83, 89 and 97.
∴ Total number of possible outcomes = 25
∴ Required probability = $\frac{25}{100}=\frac{1}{4}$
Hence, the correct answer is option (C).
Page No 161:
Question 26:
Choose the correct answer from the given four options:
A school has five houses A, B, C, D and E. A class has 23 students, 4 from house A, 8 from house B, 5 from house C, 2 from house D and rest from house E. A single student is selected at random to be the class monitor. The probability that the selected student is not from A, B and C is
(A) $\frac{4}{23}$
(B) $\frac{6}{23}$
(C) $\frac{8}{23}$
(D) $\frac{17}{23}$
Answer:
Total number of students = 23
Number of students in houses A, B and C = 4 + 8 + 5 = 17
∴ Remaining students = 23 $$ 17 = 6
So, probability that the selected student is not from houses A, B and C = $\frac{6}{23}$
Hence, the correct answer is option (B).
Page No 161:
Question 1:
The median of an ungrouped data and the median calculated when the same data is grouped are always the same. Do you think that this is a correct statement? Give reason.
Answer:
Not always, because for calculating median of a grouped data, the formula used is based on the assumption that the observations in the classes are uniformly distributed (or equally spaced).
Page No 161:
Question 2:
In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula $\overline{)x}=a+\frac{{f}_{i}{d}_{i}}{{f}_{i}}$ where a is the assumed mean. a must be one of the midpoints of the classes. Is the last statement correct? Justify your answer.
Answer:
No, it is not necessary that assumed mean consider as the mid  point of the class interval. It is considered as any value which is easy to simplify it.
Page No 162:
Question 3:
Is it true to say that the mean, mode and median of grouped data will always be different? Justify your answer.
Answer:
No, the value of these three measures can be the same, it depends on the type of data.
Page No 162:
Question 4:
Will the median class and modal class of grouped data always be different? Justify your answer.
Answer:
Not always, it depends on the given data.
Page No 162:
Question 5:
In a family having three children, there may be no girl, one girl, two girls or three girls. So, the probability of each is $\frac{1}{4}$. Is this correct? Justify your answer.
Answer:
No, the probability of each is not $\frac{1}{4}$.
Let boys be B and girls be G.
Outcomes can be {BBB, GGG, BBG, BGB, GBB, GGB, GBG, BGG}.
Total number of outcomes = 8
So,
P(No girl) = $\frac{1}{8}$
P(1 girl) = $\frac{3}{8}$
P(2 girls) = $\frac{3}{8}$
P(3 girls) = $\frac{1}{8}$
Page No 162:
Question 6:
A game consists of spinning an arrow which comes to rest pointing at one of the regions (1, 2 or 3) (Fig. 13.1). Are the outcomes 1, 2 and 3 equally likely to occur? Give reasons.
Answer:
No, the outcomes are not equally likely, because 3 contains half part of the total region, so it is more likely than 1 and 2.
Page No 162:
Question 7:
Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number 36? Why?
Answer:
Apoorv throw two dice once.
So, total number of outcomes = 6^{2} = 36
Number of outcomes for getting product 36 = 1 (6 × 6)
∴ Probability for Apoorv = $\frac{1}{36}$
Peehu throws one die.
So, total number of outcomes = 6
Number of outcomes for getting square 36 = 1 (6^{2})
∴ Probability for Peehu = $\frac{1}{6}$
Hence, Peehu has better chance of getting the number 36.
Page No 162:
Question 8:
When we toss a coin, there are two possible outcomes  Head or Tail. Therefore, the probability of each outcome is $\frac{1}{2}$. Justify your answer.
Answer:
Yes, probability of each outcome is $\frac{1}{2}$ because head and tail both are equally likely events.
Page No 162:
Question 9:
A student says that if you throw a die, it will show up 1 or not 1. Therefore, the probability of getting 1 and the probability of getting ‘not 1’ each is equal to $\frac{1}{2}$. Is this correct? Give reasons.
Answer:
No, this is not correct.
Suppose we throw a die, then total number of outcomes = 6
Possible outcomes = 1 or 2 or 3 or 4 or 5 or 6
∴ Probability of getting 1 = $\frac{1}{6}$
Now, probability of getting not 1 = 1 – Probability of getting 1
= $1\frac{1}{6}$
= $\frac{5}{6}$
Page No 162:
Question 10:
I toss three coins together. The possible outcomes are no heads, 1 head, 2 heads and 3 heads. So, I say that probability of no heads is $\frac{1}{4}$. What is wrong with this conclusion?
Answer:
Total number of outcomes = 2^{3} = 8
Possible outcomes are (HHH), (HTT), (THT), (TTH), (HHT), (THH), (HTH) and (TTT).
Now, probability of getting no head = $\frac{1}{8}$
Hence, the given conclusion is wrong because the probability of no head is $\frac{1}{8}$.
Page No 162:
Question 11:
If you toss a coin 6 times and it comes down heads on each occasion, can you say that the probability of getting a head is 1? Give reasons.
Answer:
No, if we toss a coin, then possible outcomes are head or tail.
Both are equally likely events.
So, probability is $\frac{1}{2}$.
If we toss a coin 6 times, then probability will be same in each case.
So, the probability of getting a head is not 1.
Page No 163:
Question 12:
Sushma tosses a coin 3 times and gets tail each time. Do you think that the outcome of next toss will be a tail? Give reasons.
Answer:
On tossing a coin, possible outcomes can be head or tail.
Both are equally likely events.
So, the outcome of next toss may or may not be tail.
Page No 163:
Question 13:
If I toss a coin 3 times and get head each time, should I expect a tail to have a higher chance in the 4^{th} toss? Give reason in support of your answer.
Answer:
Possible outcomes at each toss are Head or Tail.
Both are equally likely at each toss.
Hence, at the 4th toss, Tail does not have a higher chance, Head and Tail have equal chances.
Page No 163:
Question 14:
A bag contains slips numbered from 1 to 100. If Fatima chooses a slip at random from the bag, it will either be an odd number or an even number. Since this situation has only two possible outcomes, so, the probability of each is $\frac{1}{2}$. Justify.
Answer:
We know that, between 1 to 100 half numbers are even and half numbers are odd i.e. 50 numbers (2, 4, 6, 8, …, 96, 98, 100) are even and 50 numbers (1, 3, 5, 7, …, 97, 99) are odd.
So, both events are equally likely.
So, probability of getting even number = $\frac{50}{100}$ = $\frac{1}{2}$
and probability of getting odd number = $\frac{50}{100}$ = $\frac{1}{2}$
Hence, the probability of each is $\frac{1}{2}$.
Page No 166:
Question 1:
Find the mean of the distribution :
Class  1 – 3  3 – 5  5 – 7  7 – 10 
Frequency  9  22  27  17 
Answer:
We first, find the class mark x_{i} of each class and then proceed as follows.
Class  Class Marks (x_{i})  Frequency (f_{i})  f_{i} x_{i} 
1 – 3  2  9  18 
3 – 5  4  22  88 
5 – 7  6  27  162 
7 – 10  8.5  17  144.5 
$\sum _{}\left({f}_{i}\right)=75$  $\sum _{}({f}_{i}\times {x}_{i})=412.5$ 
$\mathrm{Therefore},\mathrm{mean}\left(\overline{)x}\right)=\frac{{\displaystyle \sum _{}}({f}_{i}\times {x}_{i})}{{\displaystyle \sum _{}}\left({f}_{i}\right)}=\frac{412.5}{75}=5.5\phantom{\rule{0ex}{0ex}}$
Hence, mean of the given distribution is 5.5.
Page No 166:
Question 2:
Calculate the mean of the scores of 20 students in a mathematics test :
Class  10 – 20  20 – 30  30 – 40  40 –50  50 – 60 
Number of students  2  4  7  6  1 
Answer:
We first, find the class mark x_{i} of each class and then proceed as follows.
Class  Class Marks (x_{i})  Frequency  f_{i} x_{i} 
10 – 20  15  2  30 
20 – 30  25  4  100 
30 – 40  35  7  245 
40 – 50  45  6  270 
50 – 60  55  1  55 
$\sum _{}\left({f}_{i}\right)=20$  $\sum _{}({f}_{i}\times {x}_{i})=700$ 
$\mathrm{Therefore},\mathrm{mean}\left(\overline{)x}\right)=\frac{{\displaystyle \sum _{}}({f}_{i}\times {x}_{i})}{{\displaystyle \sum _{}}\left({f}_{i}\right)}=\frac{700}{20}=35\phantom{\rule{0ex}{0ex}}$
Hence, the mean of scores of 20 students in mathematics test 35.
Page No 166:
Question 3:
Calculate the mean of the following data :
Class  4 – 7  8 – 11  12 – 15  16 – 19 
Frequency  5  4  9  10 
Answer:
Since, given data is not continuous, so we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.
Now, we first find the class mark x_{i} of each class and then proceed as follows
Class  Class Marks (x_{i})  Frequency  f_{i }x_{i} 
3.5 – 7.5  5.5  5  27.5 
7.5 – 11.5  9.5  4  38 
11.5 – 15.5  13.5  9  121.5 
15.5 – 19.5  17.5  10  175 
$\sum _{}\left({f}_{i}\right)=28$  $\sum _{}\left({f}_{i}\times {x}_{i}\right)=362$ 
$\mathrm{Therefore},\mathrm{mean}\left(\overline{)x}\right)=\frac{{\displaystyle \sum _{}}({f}_{i}\times {x}_{i})}{{\displaystyle \sum _{}}\left({f}_{i}\right)}=\frac{362}{28}=12.93\phantom{\rule{0ex}{0ex}}$
Hence, mean of the given data is 12.93.
Page No 167:
Question 4:
The following table gives the number of pages written by Sarika for completing her own book for 30 days :
Number of pages written per day  16 – 18  19 – 21  22 – 24  25 – 27  28 – 30 
Number of days  1  3  4  9  13 
Find the mean number of pages written per day.
Answer:
Since, the table is not continuous, we subtract 0.5 and 0.5 to the lower limit and upper limit respectively.
Class Mark  Midvalue (x_{i})  Number of days (f_{i})  f_{i }x_{i} 
15.5 – 18.5  17  1  17 
18.5 – 21.5  20  3  60 
21.5 – 24.5  23  4  92 
24.5 – 27.5  26  9  234 
27.5 – 30.5  29  13  377 
Total  30  780 
$\mathrm{Therefore},\mathrm{mean}\left(\overline{)x}\right)=\frac{{\displaystyle \sum _{}}({f}_{i}\times {x}_{i})}{{\displaystyle \sum _{}}\left({f}_{i}\right)}=\frac{780}{30}=26\phantom{\rule{0ex}{0ex}}$
Hence, the mean of pages written per day is 26.
Page No 167:
Question 5:
The daily income of a sample of 50 employees are tabulated as follows :
Income (in â‚ą)  1 – 200  201 – 400  401 – 600  601 – 800 
Number of employees  14  15  14  7 
Find the mean daily income of employees.
Answer:
No need to convert discontinuous classes into continuous for class mark because class mark of C.I. are same and gives same result of $\overline{x}$.
Income  x_{i}  d_{i} = (x_{i} $$ a)  f_{i}  f_{i}d_{i} 
1 – 200  100.5  $$ 200  14  $2800$ 
201 – 400  300.5  0  15  0 
401 – 600  500.5  200  14  2800 
601 – 800  700.5  400  7  2800 
$\sum _{}\left({f}_{i}\right)=50$  $\sum _{}\left({f}_{i}{d}_{i}\right)=2800$ 
∴ Assumed mean, a = 300.5 and d_{i} = (x_{i} $$ a)
$\overline{x}=a+\frac{{\displaystyle \sum _{}}\left({f}_{i}{d}_{i}\right)}{{\displaystyle \sum _{}}\left({f}_{i}\right)}$
= 300.5 + $\frac{2800}{50}$
= 356.5
Hence, the average daily income of employees is â‚ą356.5.
Page No 167:
Question 6:
An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given in the following table :
Number of seats  100 – 104  104 – 108  108 – 112  112 – 116  116 – 120 
Frequency  15  20  32  18  15 
Determine the mean number of seats occupied over the flights.
Answer:
We first, find the class mark x_{i }of each class and then proceed as follows:
Class Interval  Class Mark (x_{i})  Frequency (f_{i})  Deviation (d_{i} = x_{i} $$ a)  f_{i} d_{i} 
100 – 104  102  15  – 8  –120 
104 – 108  106  20  – 4  –80 
108 – 112  110  32  0  0 
112 – 116  114  18  4  72 
116 – 120  118  15  8  120 
N = $\sum _{}\left({f}_{i}\right)=100$  $\sum _{}({f}_{i}\times {d}_{i})=8$ 
∴ Assumed mean, a = 110, Class width, h = 4 and total observations, N = 100.
Mean $\overline{x}=a+\frac{{\displaystyle \sum _{}}({f}_{i}\times {d}_{i})}{{\displaystyle \sum _{}}\left({f}_{i}\right)}$
= 110 + $\left(\frac{8}{100}\right)$
= 110 – 0.08
= 109.92
But seats cannot be in decimal, so number of seats is 109.
Page No 167:
Question 7:
The weights (in kg) of 50 wrestlers are recorded in the following table :
Weight (in kg)  100 – 110  110 – 120  120 – 130  130 – 140  140 – 150 
Number of wrestlers  4  14  21  8  3 
Find the mean weight of the wrestlers.
Answer:
We first find the class mark x_{i}, of each class and then proceed as follows :
Weight (in kg)  Number of Wrestlers (f_{i})  Class Mark (x_{i})  Deviation (d_{i}= x_{i }– a )  f_{i }d_{i} 
100 – 110  4  105  –20  –80 
110 – 120  14  115  –10  –140 
120 – 130  21  125  0  0 
130 – 140  8  135  10  80 
140 – 150  3  145  20  60 
N = $\sum _{}\left({f}_{i}\right)=50$  $\sum _{}({f}_{i}\times {d}_{i})=80$ 
Assumed mean, (a) = 125
Class width, (h) = 10
and total observations, (N) = 50
Mean ($\overline{x}$)
= $a+\frac{{\displaystyle \sum _{}}({f}_{i}\times {d}_{i})}{{\displaystyle \sum _{}}\left({f}_{i}\right)}$
= $125+\left(\frac{80}{50}\right)$
= 125 – 1.6
= 123.4 kg
Hence, mean weight of wrestlers = 123.4 kg.
Page No 167:
Question 8:
The mileage (km per litre) of 50 cars of the same model was tested by a manufacturer and details are tabulated as given below :
Mileage (km/l)  10 – 12  12 – 14  14 – 16  16 – 18 
Number of cars  7  12  18  13 
Find the mean mileage.
The manufacturer claimed that the mileage of the model was 16 km/litre. Do you agree with this claim?
Answer:
Mileage (km/l)  Class Marks (x_{i})  Number of cars (f_{i})  f_{i}x_{i} 
10 – 12  11  7  77 
12 – 14  13  12  156 
14 – 16  15  18  270 
16 – 18  17  13  221 
Total  $\underset{}{\sum \left({f}_{i}\right)=50}$  $\sum _{}\left({f}_{i}{x}_{i}\right)=724$ 
$\mathrm{Mean}=\frac{\sum _{}^{}({f}_{i}\times {x}_{i})}{{\displaystyle \sum _{}}({f}_{i)}}$
= $\frac{724}{50}$
= 14.48
Hence, mean mileage is 14.48 km/h.
No, the manufacturer is claiming mileage 1.52 km/h more than average mileage.
Page No 168:
Question 9:
The following is the distribution of weights (in kg) of 40 persons :
Weight (in kg)  40 – 45  45 – 50  50 – 55  55 – 60  60 – 65  65 – 70  70 – 75  75 – 80 
Number of persons  4  4  13  5  6  5  2  1 
Construct a cumulative frequency distribution (of the less than type) table for the data above.
Answer:
The cumulative distribution (less than type) table is shown below :
Weight (in kg)  Cumulative Frequency 
Less than 45  4 
Less than 50  4 + 4 = 8 
Less than 55  8 + 13 = 21 
Less than 60  21 + 5 = 26 
Less than 65  26 + 6 = 32 
Less than 70  32 + 5 = 37 
Less than 75  37 + 2 = 39 
Less than 80  39 + 1 = 40 
Page No 168:
Question 10:
The following table shows the cumulative frequency distribution of marks of 800 students in an examination:
Marks  Number of students 
Below 10  10 
Below 20  50 
Below 30  130 
Below 40  270 
Below 50  440 
Below 60

570 
Below 70

670 
Below 80

740 
Below 90

780 
Below 10

800 
Construct a frequency distribution table for the data above.
Answer:
Here, we observe that 10 students have scored marks below 10 i.e. it lies between class interval 0 – 10. Similarly, 50 students have scored marks below 20. So, 50 – 10 = 40 students lie in the interval 10 – 20 and so on. The table of a frequency distribution for the given data is:
Class Interval  Number of Students 
0 – 10  10 
10 – 20  50 – 10 = 40 
20 – 30  130 – 50 = 80 
30 – 40  270 – 130 = 140 
40 – 50  440 – 270 = 170 
50 – 60  570 – 440 = 130 
60 – 70  670 – 570 = 100 
70 – 80  740 – 670 = 70 
80 – 90  780 – 740 = 40 
90 – 100  800 – 780 = 20 
Page No 169:
Question 11:
Form the frequency distribution table from the following data :
Marks (out of 90)  Number of candidates 
More than or equal to 80  4 
More than or equal to 70  6 
More than or equal to 60  11 
More than or equal to 50  17 
More than or equal to 40  23 
More than or equal to 30  27 
More than or equal to 20  30 
More than or equal to 10  32 
More than or equal to 0  34 
Answer:
Here, we observe that, all 34 students have scored marks more than or equal to 0. Since, 32 students have scored marks more than or equal to 10. So, 34 – 32 = 2 students lie in the interval 0 $$10 and so on.
Now, we construct the frequency distribution table.
Class Interval  Number of Students 
0 $$ 10  34 $$ 32 = 2 
10 $$ 20  32 $$ 30 = 2 
20 $$ 30  30 $$27 = 3 
30 $$ 40  27 $$ 23 = 4 
40 $$ 50  23 $$ 17 = 6 
50 $$ 60  17 $$ 11 = 6 
60 $$ 70  11 $$ 6 = 5 
70 $$ 80  6 $$ 4 = 2 
80 $$ 90  4 
Page No 169:
Question 12:
Find the unknown entries a, b, c, d, e, f in the following distribution of heights of students in a class :
Height (in cm)  Frequency  Cumulative frequency 
150 – 155  12  a 
155 – 160  b  25 
160 – 165  10  c 
165 – 170  d  43 
170 – 175  e  48 
175 – 180  2  f 
Total  50 
Answer:
Height (in cm)  Frequency  Cumulative frequency (given)  Cumulative Frequency 
150 – 155  12  a  12 
155 – 160  b  25  12 + b 
160 – 165  10  c  22 + b 
165 – 170  d  43  22 + b + d 
170 – 175  e  48  22 + b + d + e 
175 – 180  2  f  24 + b + d + e 
Total  50 
On comparing last two tables, we get
a = 12
∴ 12 + b = 25
⇒ b = 25$$ 12 = 13
22 + b = c
⇒ c = 22 + 13 = 35
22 + b + d = 43
⇒ 22 + 13 + d = 43
⇒ d = 43 $$ 35 = 8
and 22 + b + d + e = 48
⇒ 22 + 13 + 8 + e = 48
⇒ e = 48$$ 43 = 5
and 24 + b + d + e = f
⇒ 24 + 13 + 8 + 5 = f
∴ f = 50
The values are a = 12, b = 13, c = 35, d = 8, e = 5, f = 50.
Page No 169:
Question 13:
The following are the ages of 300 patients getting medical treatment in a hospital on a particular day :
Age (in years)  10 – 20  20 – 30  30 – 40  40 – 50  50 – 60  60 – 70 
Number of patients  60  42  55  70  53  20 
Form:
(i) Less than type cumulative frequency distribution.
(ii) More than type cumulative frequency distribution.
Answer:
(i)
We observe that the number of patients which take medical treatment in a hospital on a particular day less than 10 is 0. Similarly, less than 20 include the number of patients which take medical treatment from 010 as well as the number of patients which take medical treatment from 1020.
So, the total number of patients less than 20 is 0 + 60 = 60, we say that the cumulative frequency of the class 1020 is 60.
Less than type cumulative frequency distribution is as follows:
Age ( in years)  Number of patients 
Less than 10  0 
Less than 20  60 
Less than 30  102 
Less than 40  157 
Less than 50  227 
Less than 60  280 
Less than 70  300 
(ii)
Also, we observe that all 300 patients which take medical treatment more than or equal to 10. Since, there are 60 patients which take medical treatment in the interval 1020, this means that there are 300 – 60 = 240 patients which take medical treatment more than or equal to 20. Continuing in the same manner.
Age ( in years)  Number of patients 
More than or equal to 10  300 
More than or equal to 20  240 
More than or equal to 30  198 
More than or equal to 40  143 
More than or equal to 50  73 
More than or equal to 60  20 
More than or equal to 70  0 
Page No 170:
Question 14:
Given below is a cumulative frequency distribution showing the marks secured by 50 students of a class :
Marks  Below 20  Below 40  Below 60  Below 80  Below 100 
Number of students  17  22  29  37  50 
Form the frequency distribution table for the data.
Answer:
Here, we observe that, 17 students have scored marks below 20 i.e., it lies between class interval 020 and 22 students have scored marks below 40, so 22 – 17 = 5 students lies in the class interval 2040 continuing in the same manner, we get the complete frequency distribution table for given data.
Marks  Number of students 
0 $$ 20  17 
20 $$ 40  22 $$ 17 = 5 
40 $$ 60  29 $$ 22 = 7 
60 $$ 80  37 $$ 29 = 8 
80 $$ 100  50 $$ 37 = 13 
Page No 170:
Question 15:
Weekly income of 600 families is tabulated below :
Weekly income (in â‚ą)  Number of families 
0 – 1000  250 
1000 – 2000  190 
2000 – 3000  100 
3000 – 4000  40 
4000 – 5000  15 
5000 – 6000  5 
Total  600 
Compute the median income.
Answer:
First we construct a cumulative frequency table.
Weekly income (in â‚ą)  Number of families (f_{i})  Cumulative Frequency (c_{f}) 
0 – 1000  250  250 
1000 – 2000  190  250 + 190 = 440 
2000 – 3000  100  440 + 100 = 540 
3000 – 4000  40  540 + 40 = 580 
4000 – 5000  15  580 + 15 = 595 
5000 – 6000  5  595 + 5 = 600 
It is given that, N = 600
∴ $\frac{N}{2}$ = $\frac{600}{2}$ = 300
Since, cumulative frequency 440 lies in the interval 1000 – 2000, here, (lower median class) l = 1000, f = 190, c_{f} = 250, (class width) h = 1000 and (total observation) N = 600.
$\therefore \mathrm{Median}=l+\frac{\left({\displaystyle \frac{N}{2}}{c}_{f}\right)}{f}\times h$
= $1000+\frac{(300250)}{190}\times 1000$
= $1000+\frac{5000}{19}$
= 1000 + 263.15
= 1263.15
Hence, the median income is â‚ą1263.15.
Page No 170:
Question 16:
The maximum bowling speeds, in km per hour, of 33 players at a cricket coaching centre are given as follows :
Speed (km/h)  85 – 100  100 – 115  115 – 130  130 – 145 
Number of players  11  9  8  5 
Calculate the median bowling speed.
Answer:
First we construct the cumulative frequency table.
Speed (km/h)  Number of players  Cumulative Frequency 
85 – 100  11  11 
100 – 115  9  11 + 9 = 20 
115 – 130  8  20 + 8 = 28 
130 – 145  5  28 + 5 = 33 
It is given that, n = 33
∴ $\frac{n}{2}$ = $\frac{33}{2}$ = 16.5
So, the median class is 100 $$115, where lower limit (l) = 100
Frequency (f) = 9
Cumulative frequency (c_{f}) = 11 and class width (h) = 15
$\therefore \mathrm{Median}=l+\frac{\left({\displaystyle \frac{n}{2}}{c}_{f}\right)}{f}\times h$
= 100 + $\frac{(16.511)}{9}\times 15$
= 100 + $\frac{5.5\times 15}{9}$
= 100 + $\frac{82.5}{9}$
= 100 + 9.17
= 109.17
Hence, the median bowling speed is 109.17 km/h.
Page No 171:
Question 17:
The monthly income of 100 families are given as below :
Income (in â‚ą)  Number of families 
0 – 5000  8 
5000 – 10000  26 
10000 – 15000  41 
15000 – 20000  16 
20000 – 25000  3 
25000 – 30000  3 
30000 – 35000  2 
35000 – 40000  1 
Calculate the modal income.
Answer:
In a given data, the highest frequency is 41, which lies in the interval 10000 $$ 15000.
Here, l = 10000, f_{m} = 41, f_{1} = 26, f_{2} = 16 and h = 5000.
$\therefore \mathrm{Mode}=l+\left(\frac{{f}_{m}{f}_{1}}{2{f}_{m}{f}_{1}{f}_{2}}\right)\times h$
= 10000 + $\frac{\left(4126\right)}{(2\times 412616)}\times 5000$
= 10000 + $\left(\frac{15}{40}\right)\times 5000$
= 10000 + 15 × 125
= 10000 + 1875
= 11875
Hence, the modal income is â‚ą11875 per month.
Page No 171:
Question 18:
The weight of coffee in 70 packets are shown in the following table :
Weight (in g)  Number of packets 
200 – 201  12 
201 – 202  26 
202 – 203  20 
203 – 204  9 
204 – 205  2 
205 – 206  1 
Determine the modal weight.
Answer:
In the given data, the highest frequency is 26, which lies in the interval 201202.
Here, l = 201, f_{m} = 26, f_{1} = 12, f_{2} = 20 and (class width) h = 1.
$\therefore \mathrm{Mode}=l+\left(\frac{{f}_{m}{f}_{1}}{2{f}_{m}{f}_{1}{f}_{2}}\right)\times h$
= 201 + $\frac{\left(2612\right)}{2\times 261220}\times 1$
= 201 + $\frac{14}{20}$
= 201 + 0.7
= 201.7 g
Hence, the modal weight is 201.7 g.
Page No 171:
Question 19:
Two dice are thrown at the same time. Find the probability of getting
(i) same number on both dice.
(ii) different numbers on both dice.
Answer:
Total number of possible outcomes = 6^{2} = 36
(i) We have, same number on both dice.
So, possible outcomes are (1,1), (2,2), (3, 3), (4, 4), (5, 5) and (6, 6).
∴ Number of possible outcomes = 6
Now, required probability = $\frac{6}{36}$ = $\frac{1}{6}$
(ii) We have different number on both dice.
So, number of possible outcomes
= 36 – Number of possible outcomes for same number on both dice
= 36 $$ 6
= 30
∴ Required probability = $\frac{30}{36}$ = $\frac{5}{6}$
Page No 171:
Question 20:
Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is
(i) 7?
(ii) a prime number?
(iii) 1?
Answer:
Two dice are thrown simultaneously. [given]
So, that number of possible outcomes = 36
(i) Sum of the numbers appearing on the dice is 7.
So, the possible ways are (1, 6), (2,5), (3, 4), (4, 3), (5, 2) and (6, 1).
Number of possible ways = 6
∴ Required probability = $\frac{6}{36}$ = $\frac{1}{6}$
(ii) Sum of the numbers appearing on the dice is a prime number i.e., 2, 3, 5, 7 and 11.
So, the possible ways are (1, 1), (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6) and (6, 5).
Number of possible ways = 15
∴ Required probability = $\frac{15}{36}$ = $\frac{5}{12}$
(iii) Sum of the numbers appearing on the dice is 1.
It is not possible, so its probability is 0.
Page No 172:
Question 21:
Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is
(i) 6
(ii)12
(iii) 7
Answer:
Number of total outcomes = 36
(i) Product of the numbers on the top of the dice is 6.
So, the possible ways are (1, 6), (2, 3), (3, 2) and (6, 1).
Number of possible ways = 4
∴ Required probability = $\frac{4}{36}$ = $\frac{1}{9}$
(ii) When product of the numbers on the top of the dice is 12.
So, the possible ways are (2, 6), (3, 4), (4, 3) and (6, 2).
Number of possible ways = 4
∴ Required probability = $\frac{4}{36}$= $\frac{1}{9}$
(iii) Product of the numbers on the top of the dice cannot be 7. So, its probability is 0.
Page No 172:
Question 22:
Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is less than 9.
Answer:
Number of total outcomes = 36
When product of numbers appearing on them is less than 9, then possible ways are (1, 6), (1, 5), (1, 4), (1, 3), (1, 2), (1, 1), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (4, 2), (4, 1), (5, 1), (6, 1).
Number of possible ways = 16
Probability = $\frac{\mathrm{No}.\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{no}.\mathrm{of}\mathrm{outcomes}}$
∴ Required probability = $\frac{16}{36}$
= $\frac{4}{9}$
Page No 172:
Question 23:
Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3, respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.
Answer:
Number of total outcome = n(S) = 36
(i) Let E_{1} = Event of getting sum 2 = {(1, 1), (1, 1)}
n(E_{1}) = 2
$\therefore \mathrm{P}\left({\mathrm{E}}_{1}\right)=\frac{n\left({\mathrm{E}}_{1}\right)}{n\left(\mathrm{S}\right)}=\frac{2}{36}=\frac{1}{18}$
(ii) Let E_{2} = Event of getting sum 3 = {(1, 2), (1, 2), (2, 1), (2, 1)}
n(E_{2}) = 4
$\therefore \mathrm{P}\left({\mathrm{E}}_{2}\right)=\frac{n\left({\mathrm{E}}_{2}\right)}{n\left(\mathrm{S}\right)}=\frac{4}{36}=\frac{1}{9}$
(iii) Let E_{3} = Event of getting sum 4 = {(2, 2), (2, 2), (3, 1), (3, 1), (1, 3), (1, 3)}
n(E_{3}) = 6
$\therefore \mathrm{P}\left({\mathrm{E}}_{3}\right)=\frac{n\left({\mathrm{E}}_{3}\right)}{n\left(\mathrm{S}\right)}=\frac{6}{36}=\frac{1}{6}$
(iv) Let E_{4} = Event of getting sum 5 = {(2, 3), (2, 3), (4, 1), (4, 1), (3, 2), (3, 2)}
n(E_{4}) = 6
$\therefore \mathrm{P}\left({\mathrm{E}}_{4}\right)=\frac{n\left({\mathrm{E}}_{4}\right)}{n\left(\mathrm{S}\right)}=\frac{6}{36}=\frac{1}{6}$
(v) Let E_{5} = Event of getting sum 6 = {(3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1)}
n(E_{5}) = 6
$\therefore \mathrm{P}\left({\mathrm{E}}_{5}\right)=\frac{n\left({\mathrm{E}}_{5}\right)}{n\left(\mathrm{S}\right)}=\frac{6}{36}=\frac{1}{6}$
(vi) Let E_{6} = Event of getting sum 7 = {(4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1)}
n(E_{6}) = 6
$\therefore \mathrm{P}\left({\mathrm{E}}_{6}\right)=\frac{n\left({\mathrm{E}}_{6}\right)}{n\left(\mathrm{S}\right)}=\frac{6}{36}=\frac{1}{6}$
(vii) Let E_{7} = Event of getting sum 8 = {(5, 3), (5, 3), (6, 2), (6, 2)}
n(E_{7}) = 4
$\therefore \mathrm{P}\left({\mathrm{E}}_{7}\right)=\frac{n\left({\mathrm{E}}_{7}\right)}{n\left(\mathrm{S}\right)}=\frac{4}{36}=\frac{1}{9}$
(viii) Let E_{8} = Event of getting sum 9 = {(6, 3), (6, 3)}
n(E_{8}) = 2
$\therefore \mathrm{P}\left({\mathrm{E}}_{8}\right)=\frac{n\left({\mathrm{E}}_{8}\right)}{n\left(\mathrm{S}\right)}=\frac{2}{36}=\frac{1}{18}$
Page No 172:
Question 24:
A coin is tossed two times. Find the probability of getting at most one head.
Answer:
The possible outcomes, if a coin is tossed 2 times are S = {(HH), (TT), (HT), (TH)}
n(S) = 4
Let E = Event of getting at most one head = {(TT), (HT), (TH)}
n(E) = 3
Hence, required probability = $\frac{n\left(\mathrm{E}\right)}{n\left(\mathrm{S}\right)}=\frac{3}{4}$.
Page No 172:
Question 25:
A coin is tossed 3 times. List the possible outcomes. Find the probability of getting
(i) all heads
(ii) at least 2 heads
Answer:
The possible outcomes if a coin is tossed 3 times is
S = {(HHH), (TTT), (HTT), (THT), (TTH), (THH), (HTH), (HHT)}
n(S) = 8
(i) Let E_{1} = Event of getting all heads = {(HHH)}
∴ n(E_{1}) = 1
$\therefore \mathrm{P}\left({\mathrm{E}}_{1}\right)=\frac{n\left({\mathrm{E}}_{1}\right)}{n\left(\mathrm{S}\right)}=\frac{1}{8}$
(ii) Let E_{2} = Event of getting at least 2 heads = {(HHT), (HTH), (THH), (HHH)}
∴ n(E_{2}) = 4
$\therefore \mathrm{P}\left({\mathrm{E}}_{2}\right)=\frac{n\left({\mathrm{E}}_{2}\right)}{n\left(\mathrm{S}\right)}=\frac{4}{8}=\frac{1}{2}$
Page No 172:
Question 26:
Two dice are thrown at the same time. Determine the probability that the difference of the numbers on the two dice is 2.
Answer:
The total number of sample space in two dice, n(S) = ${6}^{2}=36$.
Let E be the event of getting the numbers whose difference is 2.
$\Rightarrow $ E = {(1, 3), (2, 4), (3, 5), (4, 6), (3, 1), (4, 2), (5, 3), (6, 4)}
∴ n(E) = 8
$\therefore \mathrm{P}\left(\mathrm{E}\right)=\frac{n\left(\mathrm{E}\right)}{n\left(\mathrm{S}\right)}=\frac{8}{36}=\frac{2}{9}$
Page No 172:
Question 27:
A bag contains 10 red, 5 blue and 7 green balls. A ball is drawn at random. Find the probability of this ball being a
(i) red ball
(ii) green ball
(iii) not a blue ball
Answer:
No. of red ball = 10
No. of blue ball = 5
No. of green balls = 7
If a ball is drawn out of 22 balls (5 blue + 7 green + 10 red), then the total number of outcomes are n(S) = 22.
(i) Let E_{1} = Event of getting a red ball
∴ n(E_{1}) = 10
$\therefore \mathrm{Required}\mathrm{probability}=\frac{n\left({\mathrm{E}}_{1}\right)}{n\left(\mathrm{S}\right)}=\frac{10}{22}=\frac{5}{11}$
(ii) Let E_{2} = Event of getting a green ball n(E_{2}) = 7.
$\therefore \mathrm{Required}\mathrm{probability}=\frac{n\left({\mathrm{E}}_{2}\right)}{n\left(\mathrm{S}\right)}=\frac{7}{22}$
(iii) Let E_{3} = Event getting a red ball or a green ball i.e. not a blue ball.
n(E_{3}) = (10 + 7) = 17
$\therefore \mathrm{Required}\mathrm{probability}=\frac{n\left({\mathrm{E}}_{3}\right)}{n\left(\mathrm{S}\right)}=\frac{17}{22}$
Page No 172:
Question 28:
The king, queen and jack of clubs are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is
(i) a heart
(ii) a king
Answer:
If we remove one king, one queen and one jack of clubs from 52 cards, then the remaining cards left, n(S) = 49
(i) Let E_{1} = Event of getting a heart
n(E_{1}) = 13
$\therefore \mathrm{Required}\mathrm{probability}=\frac{n\left({\mathrm{E}}_{1}\right)}{n\left(\mathrm{S}\right)}=\frac{13}{49}$
(ii) Let E_{2} = Event of getting a king
n(E_{2}) = 3 [Since, out of 4 kings, one club cards is already removed]
$\therefore \mathrm{Required}\mathrm{probability}=\frac{n\left({\mathrm{E}}_{2}\right)}{n\left(\mathrm{S}\right)}=\frac{3}{49}$
Page No 172:
Question 29:
The king, queen and jack of clubs are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. What is the probability that the card is
(i) a club
(ii) 10 of hearts
Answer:
i)
Total number of cards = 52 $$ 3 = 49
Let E_{1} = Event of getting a club
n(E_{1}) = (13 $$ 3) = 10
∴ Required probability = $\frac{n\left({\mathrm{E}}_{1}\right)}{n\left(\mathrm{S}\right)}$= $\frac{10}{49}$
(ii) Let E_{2} = Event of getting 10 of hearts
n(E_{2}) = 1 [In 52 playing cards only 13 are the heart cards and only one 10 in 13 heart cards]
∴ Required probability = $\frac{n\left({\mathrm{E}}_{2}\right)}{n\left(\mathrm{S}\right)}$ = $\frac{1}{49}$
Page No 172:
Question 30:
All the jacks, queens and kings are removed from a deck of 52 playing cards. The remaining cards are well shuffled and then one card is drawn at random. Giving ace a value 1 similar value for other cards, find the probability that the card has a value
(i) 7
(ii) greater than 7
(iii) less than 7
Answer:
Out of 52 playing cards, 4 jacks, 4 queens and 4 kings are removed, then the remaining cards are left, n(S) = 52 $$ 3 × 4 = 40.
(i) Let E_{1} = Event of getting a card whose value is 7
E_{1} = Card of value 7 may be of a spade, a diamond, a club or a heart
∴ n(E_{1}) = 4
$\therefore \mathrm{P}\left({\mathrm{E}}_{1}\right)=\frac{n\left({\mathrm{E}}_{1}\right)}{n\left(\mathrm{S}\right)}=\frac{4}{40}=\frac{1}{10}$
(ii) Let E_{2} = Event of getting a card whose value is greater than 7
= Event of getting a card whose value is 8, 9 or 10
∴ n(E_{2}) = 3 × 4 = 12
$\therefore \mathrm{P}\left({\mathrm{E}}_{2}\right)=\frac{n\left({\mathrm{E}}_{2}\right)}{n\left(\mathrm{S}\right)}=\frac{12}{40}=\frac{3}{10}$
(iii) Let E_{3} = Event of getting a card whose value is less than 7
= Event of getting a card whose value is 1, 2, 3, 4, 5 or 6
∴ n(E_{3}) = 6 × 4 = 24
$\therefore \mathrm{P}\left({\mathrm{E}}_{3}\right)=\frac{n\left({\mathrm{E}}_{3}\right)}{n\left(\mathrm{S}\right)}=\frac{24}{40}=\frac{3}{5}$
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Question 31:
An integer is chosen between 0 and 100. What is the probability that it is
(i) divisible by 7?
(ii) not divisible by 7?
Answer:
The number of integers between 0 and 100 is n(S) = 99
(i) Let E_{1} = Event of choosing an integer which is divisible by 7
= Event of choosing and integer which is multiple of 7
= {7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98}
∴ n(E_{1}) = 14
$\therefore \mathrm{P}\left({\mathrm{E}}_{1}\right)=\frac{n\left({\mathrm{E}}_{1}\right)}{n\left(\mathrm{S}\right)}=\frac{14}{99}$
(ii) Let Event of choosing an integer which is not divisible by 7
∴ n(E_{2}) = n(S) $$ n(E_{1})
= 99 $$14
= 85
$\therefore \mathrm{P}\left({\mathrm{E}}_{2}\right)=\frac{n\left({\mathrm{E}}_{2}\right)}{n\left(\mathrm{S}\right)}=\frac{85}{99}$
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Question 32:
Cards with numbers 2 to 101 are placed in a box. A card is selected at random. Find the probability that the card has
(i) an even number
(ii) a square number
Answer:
Total number of out comes with numbers 2 to 101, n(S) = 100
(i) Let E_{1} = Event of selecting a card which is an even number = {2, 4, 6, …100}
In an AP, l = a + (n$$1)d, here l = 100, a = 2 and d = 2
⇒ 100 = 2 + (n $$ 1)2
⇒ (n $$ 1) = 49
⇒ n = 50
n(E_{1}) = 50
∴ Required probability = $\frac{n\left({\mathrm{E}}_{1}\right)}{n\left(\mathrm{S}\right)}=\frac{50}{100}=\frac{1}{2}$
(ii) Let E_{2} = Event of selecting a card which is a square number
= {4 ,9, 16 ,25, 36, 49, 64, 81, 100}
= {(2)^{2}, (3)^{2}, (4)^{2}, (5)^{2}, (6)^{2}, (7)^{2}, (8)^{2}, (9)^{2}, (10)^{2}}
= n(E_{2}) = 9
Hence, required probability = $\frac{n\left({\mathrm{E}}_{2}\right)}{n\left(\mathrm{S}\right)}=\frac{9}{100}$
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Question 33:
A letter of English alphabets is chosen at random. Determine the probability that the letter is a consonant.
Answer:
We know that, in English alphabets, there are (5 vowels + 21 consonants) = 26 letters. So, total number of outcomes in English alphabets is n(S) = 26
Let E = Event of choosing an English alphabet, which is a consonant
= {b, c, d, f, g, h, j, k, l, m, n, p, q, r, s t, v, w, x, y, z}
∴ n(E) = 21
Hence, required probability $=\frac{n\left(\mathrm{E}\right)}{n\left(\mathrm{S}\right)}=\frac{21}{26}$
Page No 173:
Question 34:
There are 1000 sealed envelopes in a box, 10 of them contain a cash prize of Rs 100 each, 100 of them contain a cash prize of Rs 50 each and 200 of them contain a cash prize of Rs 10 each and rest do not contain any cash prize. If they are well shuffled and an envelope is picked up out, what is the probability that it contains no cash prize?
Answer:
Total number of sealed envelopes in a box, n(S) = 1000
Number of envelopes containing cash prize = 10 + 100 + 200 = 310
Number of envelopes containing no cash prize,
n(E) = 1000 $$ 310 = 690
P(E) = $\frac{n\left(\mathrm{E}\right)}{n\left(\mathrm{S}\right)}=\frac{690}{1000}=\frac{69}{100}$
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Question 35:
Box A contains 25 slips of which 19 are marked Re 1 and other are marked Rs 5 each. Box B contains 50 slips of which 45 are marked Re 1 each and others are marked Rs 13 each. Slips of both boxes are poured into a third box and reshuffled. A slip is drawn at random. What is the probability that it is marked other than Re 1?
Answer:
Total number of slips in a box, n(S) = 25 + 50 = 75
From the chart it is clear that, there are 11 slips which are marked other than Rs 1.
∴ Required probability = $\frac{\mathrm{No}.\mathrm{of}\mathrm{slips}\mathrm{other}\mathrm{than}\mathrm{Re}.1}{\mathrm{Total}\mathrm{no}.\mathrm{of}\mathrm{slips}}=\frac{11}{75}$
Page No 173:
Question 36:
A carton of 24 bulbs contain 6 defective bulbs. One bulbs is drawn at random. What is the probability that the bulb is not defective? If the bulb selected is defective and it is not replaced and a second bulb is selected at random from the rest, what is the probability that the second bulb is defective?
Answer:
Total number of bulbs, n(S) = 24
Let E_{1} = Event of selecting not defective bulb = Event of selecting good bulbs
n(E_{1}) = 18
$\therefore \mathrm{P}\left(\frac{{\mathrm{E}}_{1}}{\mathrm{S}}\right)=\frac{18}{24}=\frac{3}{4}$
Suppose, the selected bulb is defective and not replaced, then total number of bulbs remains in a carton, n(S) = 23
In them, 18 are good bulbs and 5 are defective bulbs.
∴ P(selecting second defective bulb) = $\frac{5}{23}$
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Question 37:
A child’s game has 8 triangles of which 3 are blue and rest are red, and 10 squares of which 6 are blue and rest are red. One piece is lost at random. Find the probability that it is a
(i) triangle
(ii) square
(iii) square of blue colour
(iv) triangle of red colour
Answer:
Total number of figures
n(S) = 8 triangles + 10 squares = 18
(i) P(lost piece is a triangle) = $\frac{8}{18}$ = $\frac{4}{9}$
(ii) P(lost piece is a square) = $\frac{10}{18}$ = $\frac{5}{9}$
(iii) P(square of blue colour) = $\frac{6}{18}$ = $\frac{1}{3}$
(iv) P(triangle of red colour) = $\frac{5}{18}$
Page No 173:
Question 38:
In a game, the entry fee is Rs 5. The game consists of a tossing a coin 3 times. If one or two heads show, Sweta gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise she will lose. For tossing a coin three times, find the probability that she
(i) loses the entry fee.
(ii) gets double entry fee.
(iii) just gets her entry fee.
Answer:
Total possible outcomes of tossing a coin 3 times,
S = {(HHH), (TTT), (HTT), (THT), (TTH), (THH), (HTH), (HHT)}
∴ n(S) = 8
(i) Let E_{1} = Event that Sweta losses the entry fee
= She tosses tail on three times
n(E_{1}) = {(T T T)}
$\mathrm{P}\left({\mathrm{E}}_{1}\right)=\frac{n\left({\mathrm{E}}_{1}\right)}{n\left(\mathrm{S}\right)}=\frac{1}{8}$
(ii) Let Event that Sweta gets double entry fee
= She tosses tail on three times = {(HHH)}
n(E_{2}) = 1
$\mathrm{P}\left({\mathrm{E}}_{2}\right)=\frac{n\left({\mathrm{E}}_{2}\right)}{n\left(\mathrm{S}\right)}=\frac{1}{8}$
(iii) Let event that Sweta gets her entry fee back
= Sweta gets heads one or two times
= {(HTT), (THT), (TTH), (HHT), (HTH), (THH)}
n(E_{3}) = 6
$\mathrm{P}\left({\mathrm{E}}_{3}\right)=\frac{n\left({\mathrm{E}}_{3}\right)}{n\left(\mathrm{S}\right)}=\frac{6}{8}=\frac{3}{4}$
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Question 39:
A die has its six faces marked 0, 1, 1, 1, 6, 6. Two such dice are thrown together and the total score is recorded.
(i) How many different scores are possible?
(ii) What is the probability of getting a total of 7?
Answer:
Given, a die has its six faces marked {0, 1, 1, 1, 6, 6}
∴ Total sample space, n(S) = 6^{2} = 36
(i)Number of favorable outcomes are (0, 0), (0, 1), (0, 6), (1, 0), (1, 1), (1, 6), (6, 0), (6, 1), (6, 6).
The different score which are possible are 6 scores i.e. 0, 1, 2, 6, 7 and 12.
(ii) Let E = Event of getting a sum 7
= {(1, 6), (1, 6), (1, 6), (1, 6), (1, 6), (1, 6), (6, 1), (6, 1), (6, 1), (6, 1), (6, 1), (6, 1)}
∴ n(E) = 12
P(E) = $\frac{n\left(\mathrm{E}\right)}{n\left(\mathrm{S}\right)}=\frac{12}{36}=\frac{1}{3}$
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Question 40:
A lot consists of 48 mobile phones of which 42 are good, 3 have only minor defects and 3 have major defects. Varnika will buy a phone if it is good but the trader will
only buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is
(i) acceptable to Varnika?
(ii) acceptable to the trader?
Answer:
Given, total number of mobile phones
n(S) = 48
(i) Let E_{1} = Event that Varnika will buy a mobile phone
= Varnika buy only, if it is good mobile
∴ n(E_{1}) = 42
$\therefore P\left({E}_{1}\right)=\frac{n\left({E}_{1}\right)}{n\left(S\right)}=\frac{42}{48}=\frac{7}{8}$
(ii) Let E_{2} = Event that trader will buy only when it has no major defects
= Trader will buy only 45 mobiles
∴ n(E_{2}) = 45
$\therefore P\left({E}_{2}\right)=\frac{n\left({E}_{2}\right)}{n\left(S\right)}=\frac{45}{48}=\frac{15}{16}$
Page No 174:
Question 41:
A bag contains 24 balls of which x are red, 2x are white and 3x are blue. A ball is selected at random. What is the probability that it is
(i) not red?
(ii) white?
Answer:
Given that,
A bag contains 24 balls of which x are red, 2x are white and 3x are blue.
Total number of balls = x + 2x + 3x =24
⇒ 6x = 24
⇒ x = 4
∴ Number of red balls = x = 4
Number of white balls = 2x = 2×4 = 8
and number of blue balls = 3x = 3×4 = 12
So, total number of outcomes for a ball is selected at random in a bag contains 24 balls.
⇒ n(S) = 24
(i) Let E_{1} = Event of selecting a ball which is not red i.e.can be white or blue.
∴ n(E_{1}) =Number of white balls + Number of blue balls
⇒ ∴ n(E_{1}) = 8 + 12 = 20
$\therefore \mathrm{Required}\mathrm{Probability}=\frac{\mathrm{n}\left({\mathrm{E}}_{1}\right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{20}{24}=\frac{5}{6}$
(ii) Let E_{2} Event of selecting a ball which is white.
⇒ ∴ n(E_{2}) = 8
So, required probability = $\frac{\mathrm{n}\left({\mathrm{E}}_{2}\right)}{\mathrm{n}\left(\mathrm{S}\right)}=\frac{8}{24}=\frac{1}{3}$
Page No 174:
Question 42:
At a fete, cards bearing numbers 1 to 1000, one number on one card, are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square greater than 500, the player wins a prize. What is the probability that
(i) the first player wins a prize?
(ii) the second player wins a prize, if the first has won?
Answer:
Given that, at a fete, cards bearing numbers 1 to 1000 one number on one card, are put in a box. Each player selects one card at random and that card is not replaced so, the total number of outcomes are n(S) = 1000
If the selected card has a perfect square greater than 500, then player wins a prize.
(i) Let E_{1} = Event first player wins a prize = Player select a card which is a perfect square greater than 500
= {529, 576, 625, 676, 729, 784, 841, 900, 961}
= {(23)^{2}, (24)^{2}, (25)^{2}, (26)^{2}, (27)^{2}, (28)^{2}, (29)^{2}, (30)^{2}, (31)^{2}}
∴ n(E) = 9
So, required probability = $\frac{n\left({E}_{1}\right)}{n\left(S\right)}=\frac{9}{1000}=0.009$
(ii) First, has won i.e., one card is already selected, greater than 500, has a perfect square. Since, repetition is not allowed. So, one card is removed out of 1000 cards. So, number of remaining card is 999.
∴ Total number of remaining outcomes, n(S’) = 999
Let E_{2} be the event that the second player wins a prize, if the first has won.
Then, the remaining cards has a perfect square greater than 500 = 8
$\therefore n\left({E}_{2}\right)=91=8$
So, required probability = $\frac{n\left({E}_{2}\right)}{n(S\text{'})}=\frac{8}{999}$
View NCERT Solutions for all chapters of Class 10