RS Aggarwal 2018 Solutions for Class 8 Math Chapter 25 - Probability

Answer:

(i) The possible outcomes are head (H) and tail (T).

(ii) The possible outcomes are HH, HT, TH and TT.

(iii) The possible outcomes are 1, 2, 3, 4, 5 and 6.

(iv) The total number of possible outcomes is 52.

Answer:

The possible outcomes in a coin toss are H and T.
Total number of outcomes = 2
Number of tails = 1
∴ ${\mathrm{P}}_{\left(\mathrm{tail}\right)}=\frac{1}{2}$

Answer:

The outcomes when two coins are tossed are HH, HT, TH and TT.
i.e., total no. of possible outcomes = 4

(i) Getting both tails means TT.
Number of outcomes with two tails = 1
∴ ${\mathrm{P}}_{\left(\mathrm{both} \mathrm{tails}\right)}=\frac{1}{4}$

Answer:

Total number of balls $= 4+5=9$
(i) Number of white balls = 4
∴​ ${\mathrm{P}}_{\left(\mathrm{white} \mathrm{ball}\right)}=\frac{4}{9}$

Number of blue balls = 5
∴​ ${\mathrm{P}}_{\left(\mathrm{blue} \mathrm{ball}\right)}=\frac{5}{9}$

Answer:

Total number of balls $=5+6+4=15$
(i) Number of green balls = 4
∴​ ${\mathrm{P}}_{\left(\mathrm{green} \mathrm{ball}\right)}=\frac{4}{15}$

(ii) Number of white balls = 5
∴​ ${\mathrm{P}}_{\left(\mathrm{white} \mathrm{ball}\right)}=\frac{5}{15}=\frac{1}{3}$

(iii) Number of balls that are not red (i.e., 5 white and 4 green)$= 9$
∴​ ${\mathrm{P}}_{\left(\mathrm{non}-\mathrm{red} \mathrm{balls}\right)}=\frac{9}{15}=\frac{3}{5}$

Answer:

Total number of tickets $=10+20=30$

Number of prize tickets = 10

∴​ ${\mathrm{P}}_{\left(\mathrm{getting} \mathrm{a} \mathrm{prize}\right)}=\frac{10}{30}=\frac{1}{3}$

Answer:

Total number of bulbs in the box = 100
(i) Number of defective bulbs = 8
∴​ ${\mathrm{P}}_{\left(\mathrm{defective} \mathrm{bulb}\right)}=\frac{8}{100}=\frac{2}{25}$

(ii) Number of functioning bulbs= $100-8=92$
∴​ ${\mathrm{P}}_{\left(\mathrm{non}-\mathrm{defective} \mathrm{bulb}\right)}=\frac{92}{100}=\frac{23}{25}$

Answer:

The possible outcomes when a dice is thrown at random are 1, 2, 3, 4, 5 and 6.
Total number of outcomes = 6

(i) ∴​ ${\mathrm{P}}_{\left(\mathrm{getting} 2\right)}=\frac{1}{6}$

(ii) The numbers less than 3 are 1 and 2.
      Number of possible outcomes = 2
      ∴​ ${\mathrm{P}}_{\left(\mathrm{getting} \mathrm{a} \mathrm{number}<3\right)}=\frac{2}{6}=\frac{1}{3}$

(iii) A composite number is defined as a number with at least one positive divisor other than itself and unity.
In a dice, 4 and 6 are composite numbers.
Number of possible outcomes = 2
∴​ ${\mathrm{P}}_{\left(\mathrm{getting} \mathrm{a} \mathrm{composite} \mathrm{number}\right)}=\frac{2}{6}=\frac{1}{3}$

Answer:

Total number of ladies surveyed $=200$

Number of ladies who dislike coffee $=118$

If chosen randomly, ${\mathrm{P}}_{(\mathrm{a} \mathrm{lady} \mathrm{that} \mathrm{dislikes} \mathrm{coffee})} = \frac{118}{200}=\frac{59}{100}$

Answer:

Total number of possible outcomes$=19$
(i) The prime numbers between 1 and 19 are 2, 3, 5, 7, 11, 13, 17 and 19.
     Total number of primes $=8$   
     ∴ ${\mathrm{P}}_{\left(\mathrm{prime} \mathrm{number}\right)}=\frac{8}{19}$

Answer:

Total number of possible outcomes$=52$
(i) There are 4 kings cards (king of hearts, king of diamonds, king of spades and king of cloves)
Number of kings$=4$
∴ ${\mathrm{P}}_{\left(\mathrm{king}\right)}=\frac{4}{52}=\frac{1}{13}$

(ii) There is a total of 13 spades cards.
Number of spades$=13$
∴ ${\mathrm{P}}_{\left(\mathrm{spades}\right)}=\frac{13}{52}=\frac{1}{4}$
(iii) There are 2 red queens in a pack (queen of hearts and queen of diamonds)
Number of red queens $=2$
∴ ${\mathrm{P}}_{\left(\mathrm{red} \mathrm{queen}\right)}=\frac{2}{52}=\frac{1}{26}$

Answer:

Total number of possible outcomes $=52$
(i) There are 4 cards of with the number 4 (4 of hearts, 4 of diamonds, 4 of spades and 4 of cloves)
∴ ${\mathrm{P}}_{\left(4 \mathrm{card}\right)}=\frac{4}{52}=\frac{1}{13}$
(ii) There are 4 queens in a pack of cards (queen of hearts, queen of diamonds, queen of spades and queen of cloves)
∴ ${\mathrm{P}}_{\left(\mathrm{queen}\right)}=\frac{4}{52}=\frac{1}{13}$
(iii) There are a total of 26 black cards (13 spade cards and 13 clove cards)
∴ ${\mathrm{P}}_{\left(\mathrm{black} \mathrm{card}\right)}=\frac{26}{52}=\frac{1}{2}$

Answer:

(b) $\frac{5}{8}$

The wheel has a total of $5+3=8$ sectors.
Number of green sectors = 5
Now, ${\mathrm{P}}_{\left(\mathrm{getting} \mathrm{a} \mathrm{green} \mathrm{sector}\right)}=\frac{5}{8}$

Answer:

(c) $\frac{3}{8}$

Total number of cards $=8$
Number of cards with numbers less than 4 $=3$ (cards with numbers 1, 2 and 3)
Now, ${\mathrm{P}}_{\left(\mathrm{getting} \mathrm{a} \mathrm{number} \mathrm{less} \mathrm{than} 4\right)}=\frac{3}{8}$

Answer:

(b) $\frac{1}{2}$

When two coins are tossed, the possible outcomes are HH, HT, TH and TT.
​Total number of outcomes $=4$
Number of outcomes with one head and one tail $=2$
$\mathrm{Now}, {\mathrm{P}}_{\left(\mathrm{one} \mathrm{head} \mathrm{and} \mathrm{one} \mathrm{tail}\right)}=\frac{2}{4}=\frac{1}{2}$

Answer:

(d) $\frac{2}{5}$

Total number of outcomes $=5$
Number of red balls $=2$
$\mathrm{Now}, {\mathrm{P}}_{\left(\mathrm{red} \mathrm{ball}\right)}=\frac{2}{5}$

Answer:

(b) $\frac{1}{6}$

The possible outcomes are 1, 2, 3, 4, 5 and 6.
Total number of outcomes $=6$
Now, ${\mathrm{P}}_{(\mathrm{getting} 6)} = \frac{1}{6}$

Answer:

(a) $\frac{1}{2}$

Total number of outcomes $=6$ (Numbers: 1, 2, 3, 4, 5 and 6)
The even numbers are 2, 4, and 6.
Number of favourable outcomes $=3$
Now, ${\mathrm{P}}_{\left(\mathrm{even} \mathrm{number}\right)}=\frac{3}{6}=\frac{1}{2}$

Answer:

(c) $\frac{1}{13}$

Total number of cards $=52$
Number of queens = 4 (i.e., queen of hearts, queen of diamonds, queen of cloves and queen of spades)
$\mathrm{Now}, {\mathrm{P}}_{\left(\mathrm{queen}\right)}=\frac{4}{52}=\frac{1}{13}$

Answer:

(b) $\frac{1}{26}$