Probability(Quantitative Aptitude ) Questions and answers for exam Preparation

Quantitative Aptitude

PROBABILITY MCQs

Probability, Probability I

Total Questions : 775 | Page 12 of 78 pages

Question 111. Three children are selected at random from a group of 6 boys and 4 girls. It is known that in this group exactly one girl and one boy belong to same parent. The probability that the selected group of children have no blood relations, is equal to

115
1315
1415
215

Discuss Question

Answer: Option C. -> 1415
:
C
P(boyand his sister both are selected) =

\frac{8 C_{1}}{10 C_{3}} = \frac{1}{15}

∴

required probability =

1 - \frac{1}{15} = \frac{14}{15}

Question 112. Two players A and B throw a die alternately for a prize of Rs 11, which is to be won by a player who first throws a six. If A starts the game, their respective expectations are

Rs 6; Rs 5
Rs 7; Rs 4
Rs 5.50; Rs 5.50
Rs 5.75; Rs 5.25

Discuss Question

Answer: Option A. -> Rs 6; Rs 5
:
A
Probability through a six =

\frac{1}{6}

P (A) = \frac{1}{6}, P (¯ A) = \frac{5}{6}, P (B) = \frac{1}{6}, P (¯ B) = \frac{5}{6}

Probability of A winning

= P (A) + P (¯ A) P (¯ B) P (A) + P (¯ A) P (¯ B) P (¯ A) P (¯ B) P (A) + . . .

= \frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} + . . .

= \frac{\frac{1}{6}}{1 - \frac{25}{36}} = \frac{6}{11}

Probability of B winning

= 1 - \frac{6}{11} = \frac{5}{11}

∴

Expectations of A and B are

\frac{6}{11} \times 11 =

Rs 6 and

\frac{5}{11} \times 11

=Rs 5

Question 113. Let A, B , and C be three independent events with

P (A) = \frac{1}{3}, P (B) = \frac{1}{2}, a n d P (C) = \frac{1}{4} .

. The probability of exactly 2 of these events occurring, is equal to

14
724
34
1724

Discuss Question

Answer: Option A. -> 14
:
A
P(exactly two of A, B and C) = P(AB) + P(BC) + P(CA) – 3 P(ABC)
Since A, B ,C are independent events,
Required probability = P(A)P(B) + P(B)P(C) +P(C)P(A) – 3 P(A)P(B)P(C)

= \frac{1}{3} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{4} + \frac{1}{4} \times \frac{1}{3} - 3 \frac{1}{3} \times \frac{1}{2} \times \frac{1}{4} = \frac{1}{4.}

Question 114. A box contains 24 identical balls of which 12 are white and 12 black. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the

4^{t h}

time on the

7^{t h}

draw is

564
2732
532
12

Discuss Question

Answer: Option C. -> 532
:
C
In any trail, P(getting white ball) =

\frac{1}{2}

P(getting black ball) =

\frac{1}{2}

Now, required event will occur if in the first six trails 3 white balls are drawn in any one of the 3 trails from six. The remaining 3 trails must be kept reserved for black balls. This can happen
In

^{6} C_{3} \times^{3} C_{3} = 20

ways.

= 20 \times {(\frac{1}{2})}^{3} \times {(\frac{1}{2})}^{3} \times \frac{1}{2} = \frac{5}{32}

Question 115. Sixteen players

P_{1} . P_{2} . \dots \dots \dots P_{16}

play in a tournament. They are divided into eight pairs at random.From each pair a winner is decided on the basis of a game played between the two players of the pair.Assuming that all the players are of equal strength, the probability that exactly one of the two players

P_{1}

and

P_{2}

is among the eight winners is

415
715
815
1730

Discuss Question

Answer: Option C. -> 815
:
C
Let

E_{1} (E_{2})

denote the event that

P_{1}

and

P_{2}

are paired (not paired) together and let A denote the event that one of two players

P_{1}

and

P_{2}

is among the winners.
Since, P1 can be paired with any of the remaining 15 players.
We have,

P (E_{1}) = \frac{1}{15}

and

P (E_{2}) = 1 - P (E_{1}) = 1 - \frac{1}{15} = \frac{14}{15}

In case

E_{1}

occurs, it is certain that one of

P_{1}

and

P_{2}

will be among the winners. In case

E_{2}

occurs, the probability that exactly of

P_{1}

and

P_{2}

is among the winners is

P {(P_{1} \cap_{2}) \cup (_{1} \cap P_{2})} = P (P_{1} \cap_{2}) + P (_{1} \cap P_{2}) = P (P_{1}) P (_{2}) + P (_{1}) P (P_{2}) = (\frac{1}{2}) (1 - \frac{1}{2}) + (1 - \frac{1}{2}) (\frac{1}{2}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}

ie,

P (\frac{A}{E_{1}}) = 1

and

P (\frac{A}{E_{2}}) = \frac{1}{2}

By the total probability Rule,

P (A) = P (E_{1}) . P (\frac{A}{E_{1}}) + P (E_{1}) . P (\frac{A}{E_{2}}) = \frac{1}{15} (1) + \frac{14}{15} (\frac{1}{2}) = \frac{8}{15}

Question 116. In four schools

B_{1}, B_{2}, B_{3}, B_{4}

the percentage of girls students is 12, 20, 13, 17 respectively. From a school selected at random, one student is picked up at random and it is found that the student is a girl. The probability that the school selected is

B_{2}

, is

631
1031
1362
1762

Discuss Question

Answer: Option B. -> 1031
:
B
Favorable number of cases =

^{20} C_{1} = 20

Sample space =

^{62} C_{1} = 62

∴

Required probability =

\frac{20}{62} = \frac{10}{31}

Question 117. The probability that A speaks truth is

\frac{4}{5}

, while this probability for B is

\frac{3}{4}

. The probability that they contradict each other when asked to speak on a fact is

323
120
15
1440

Discuss Question

Answer: Option D. -> 1440
:
D
In short the event described here is P(A)'P(B) +P(A)P(B)'
where P(A) and P(B) arethe probabilitiesof A and B speaking the truth respectively.
P(E)= P(A)'P(B) +P(A)P(B)' =

\frac{1}{5} \times \frac{3}{4} + \frac{4}{5} \times \frac{1}{4}

P(E)=

\frac{3}{20} + \frac{4}{20} = \frac{7}{20} = \frac{14}{40}

Question 118. Ashu studies at Byju's classes and her probability of selection in IIT-JEE is

\frac{4}{5}

. Ridhima took coaching at FIIT-JEE and the probability of her selection is

\frac{2}{3}

. What is the probability that only 1 of them cracks the Exam?

25
310
415
None of the above.

Discuss Question

Answer: Option A. -> 25
:
A
P(A)=

\frac{4}{5}

P(R)=

\frac{2}{3}

P(E)= P(A'R)+P(AR')
P(E)=

\frac{1}{5} \times \frac{2}{3} + \frac{4}{5} \times \frac{1}{3}

P(E)=

\frac{2}{15} + \frac{4}{15}

P(E)=

\frac{6}{15}

P(E)=

\frac{2}{5}

Question 119. Let X be a set containing n elements. If two subsets A and B of X are picked at random, the probability that A and B the same number of elements, is

2nCn22n
12nCn
1.3.5....(2n−1)2n
3n4n

Discuss Question

Answer: Option A. -> 2nCn22n
:
A
We know that the number of sub-sets of a set containing n elements is

2^{n}

.
Therefore the number of ways of choosing A and B is

2^{n} . 2^{n} = 2^{2 n}

We also know that the number of sub-sets (of X) which contain exactly r elements is

^{n} C_{r} .

Therefore the number of ways of choosing A and B, so that they have the same number elements is

(^{n} C_{0})^{2} + (^{n} C_{1})^{2} + (^{n} C_{2})^{2} + . . . . + (^{n} C_{n})^{2} =^{2 n} C_{n}

Thus the required probability =

\frac{^{2 n} C_{n}}{2^{2 n}} .

Question 120. Six boys and six girls sit in a row. What is the probability that the boys and girls sit alternatively

1462
1924
12
None of these

Discuss Question

Answer: Option A. -> 1462
:
A
Let n = total number of ways = 12!
and m = favourable numbers of ways =

2 \times 6! . 6!

Since the boys and girls can sit alternately in 6! . 6! ways if we begin with a boy and similarly
they can sit alternately in 6! . 6! Ways if we begin with a girl
Hence required probability =