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61. A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?

A. $\frac{3}{4}$

B. $\frac{4}{7}$

C. $\frac{1}{8}$

D. $\frac{3}{7}$

Discuss

Solution:

Total number of balls = (6 + 8) = 14
Number of white balls = 8
P (drawing a white ball) =

\frac{8}{14}

\frac{4}{7}

62. A bag contains 4 red, 5 yellow and 6 pink balls. Two balls are drawn at random. What is the probability that none of the balls drawn are yellow?

A. $\frac{1}{7}$

B. $\frac{3}{7}$

C. $\frac{2}{7}$

D. $\frac{5}{14}$

Discuss

Solution:

Number of red balls = 4
Number of yellow balls = 5
Number of pink balls = 6
Total balls = 4 + 5 + 6 = 15
Total possible outcomes = selection of 2 balls out of 15 balls
=

^{15} C_{2}

\frac{15!}{2! (15 - 2)!}

\frac{15!}{2! \times 13!}

\frac{15 \times 14}{1 \times 2}

=105
Total favourable outcomes = selection of 2 balls out of 4 orange and 6 pink balls.

=^{10} C_{2}

\frac{10!}{2! (10 - 2)!}

\frac{10!}{2! \times 8!}

\frac{10 \times 9}{1 \times 2}

= 45
∴ Required probability =

\frac{45}{105}

\frac{3}{7}

63. In a simultaneous throw of two dice, what is the probability of getting a doublet ?

A. $\frac{1}{6}$

B. $\frac{1}{4}$

C. $\frac{2}{3}$

D. $\frac{3}{7}$

Discuss

Solution:

In a simultaneous throw of dice, n (S) = (6 × 6) = 36
Let E = event of getting a doublet
= [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)]
∴ P(E) =

\frac{n (E)}{n (S)}

\frac{6}{36}

\frac{1}{6}

64. An urn contains 6 red, 5 blue and 2 green marbles. If 2 marbles are picked at random, what is the probability that both are red?

A. $\frac{6}{13}$

B. $\frac{5}{26}$

C. $\frac{5}{13}$

D. $\frac{7}{26}$

Discuss

Solution:

P(Both are red),

\begin{aligned} = \frac{^{6} C_{2}}{^{13} C_{2}} \\ = \frac{5}{26} \end{aligned}

65. You toss a coin AND roll a die. What is the probability of getting a tail and a 4 on the die?

A. $\frac{1}{2}$

B. $\frac{1}{6}$

C. $\frac{1}{8}$

D. $\frac{1}{12}$

Discuss

Solution:

Probability of getting a tail when a single coin is tossed

= \frac{1}{2}

Probability of getting 4 when a die is thrown

= \frac{1}{6}

Required probability,

\begin{aligned} = \frac{1}{2} \times \frac{1}{6} \\ = \frac{1}{12} \end{aligned}

66. Two cards are drawn from a pack of well shuffled cards. Find the probability that one is a club and other in King.

A. $\frac{1}{13}$

B. $\frac{4}{13}$

C. $\frac{1}{52}$

D. $\frac{1}{26}$

Discuss

Solution:

Let X be the event that cards are in a club which is not king and other is the king of club.
Let Y be the event that one is any club card and other is a non-club king.
Hence, required probability:

\begin{aligned} = P (A) + P (B) \\ = \frac{^{12} C_{1} \times^{1} C_{1}}{^{52} C_{2}} + \frac{^{13} C_{1} \times^{3} C_{1}}{^{52} C_{2}} \\ = (\frac{2 \times (12 \times 1)}{52 \times 51}) + (\frac{2 (13 \times 3)}{52 \times 51}) \\ = (\frac{24 + 78}{52 \times 51}) \\ = \frac{1}{26} \end{aligned}

67. P and Q sit in a ring arrangement with 10 persons. What is the probability that P and Q will sit together?

A. $\frac{2}{11}$

B. $\frac{3}{11}$

C. $\frac{2}{9}$

D. $\frac{4}{9}$

Discuss

Solution:

n(S)= number of ways of sitting 12 persons at round table:
= (12 - 1)! = 11!
Since two persons will be always together, then number of persons:
= 10 + 1 = 11
So, 11 persons will be seated in (11 - 1)! = 10! ways at round table and 2 particular persons will be seated in 2! ways.
n(A) = The number of ways in which two persons always sit together = 10! × 2

\begin{aligned} P (A) = \frac{n (A)}{n (S)} \\ = \frac{10! \times 2!}{11!} \\ = \frac{2}{11} \end{aligned}

68. In a race, the odd favour of cars P, Q, R, S are 1 : 3, 1 : 4, 1 : 5 and 1 : 6 respectively. Find the probability that one of them wins the race.

A. $\frac{9}{17}$

B. $\frac{114}{121}$

C. $\frac{319}{420}$

D. $\frac{27}{111}$

Discuss

Solution:

Let the probability of winning the race is denoted by P(person)

\begin{aligned} P (P) = \frac{1}{4}, \\ P (Q) = \frac{1}{5}, \\ P (R) = \frac{1}{6}, \\ P (S) = \frac{1}{7} \end{aligned}

All the events are mutually exclusive (since if one of them wins then other would lose as pointed out by rahul) hence,
Required probability:

= P (P) + P (Q) + P (R)

+ P (S)

\begin{aligned} = \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} \\ = \frac{319}{420} \end{aligned}

69. A card is drawn from a pack of 52 cards. The card is drawn at random. What is the probability that it is neither a spade nor a Jack?

A. $\frac{4}{13}$

B. $\frac{2}{13}$

C. $\frac{6}{13}$

D. $\frac{9}{13}$

Discuss

Solution:

There are 13 spade and 3 more jack
Probability of getting spade or a jack:

\begin{aligned} = \frac{13 + 3}{52} \\ = \frac{4}{13} \end{aligned}

So probability of getting neither spade nor a jack:

\begin{aligned} = 1 - \frac{4}{13} \\ = \frac{9}{13} \end{aligned}

70. What is the probability that a number selected from numbers 1, 2, 3, ......, 30, is prime number, when each of the given numbers is equally likely to be selected?

A. $\frac{9}{30}$

B. $\frac{8}{30}$

C. $\frac{10}{30}$

D. $\frac{11}{30}$

Discuss

Solution:

X = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}
n(X) = 10, n(S) = 30
Hence required probability,

\begin{aligned} = \frac{n (X)}{n (S)} \\ = \frac{10}{30} \end{aligned}

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