Probability(Quantitative Aptitude ) Questions and answers for exam Preparation

Quantitative Aptitude

PROBABILITY MCQs

Probability, Probability I

Total Questions : 775 | Page 14 of 78 pages

Question 131. Urn A contains 6 red , 4 white balls and urn B contains 4 red and 6 white balls. One ball is drawn from the urn A and placed in the urn B. Then one ball is drawn at random from urn B and placed in the urn A. if one ball is now drawn from the urn A, then the probability that it is found to be red is

3255
3040
3250
34

Discuss Question

Answer: Option A. -> 3255
:
A
case(i): red ball from A to B, red ball from B to A ,then red ball from A

P_{1} = \frac{6}{10} \times \frac{5}{11} \times \frac{6}{10}

Case(ii): red ball from A to B, white ball from B to A, red ball from A

P_{2} = \frac{6}{10} \times \frac{6}{11} \times \frac{5}{10}

Case(iii): white ball from A to B, red ball from B to A, red ball from A

P_{3} = \frac{4}{10} \times \frac{4}{11} \times \frac{7}{10}

Case(iv): white ball from A to B, white ball from B to A, red ball from A

P_{4} = \frac{4}{10} \times \frac{7}{11} \times \frac{6}{10}

Therefore required probability

= P_{1} + P_{2} + P_{3} + P_{4} = \frac{640}{1100} = \frac{32}{55}

Question 132. 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 133. In a multiple choice question there are four alternative answers of which one or more than one is or are correct. A candidate will get marks on the question only if he ticks all correct answers. The candidate decides to tick answers at random. If he is allowed up to three chances to answer the question, the probability that he will get marks on it is given by

1−(1415)3
(115)3
15
1415

Discuss Question

Answer: Option C. -> 15
:
C
The total number of ways of answering one or more alternatives of 4 is

4 C_{1} + 4 C_{2} + 4 C_{3} + 4 C_{4} = 15,

Out of these 15 combinations, only one combination is correct. The probability of answeringthe alternaive correctly at the first trial is

\frac{1}{15},

that at the second trial is

(\frac{14}{15} . \frac{1}{14}) = \frac{1}{15,}

and that at the third trail is

\frac{14}{15} . \frac{13}{14} . \frac{1}{13} = \frac{1}{15},

Therefore the probability that the andidate will get marks on the question if he allowed upto three chances, is

\frac{1}{15} + \frac{1}{15} + \frac{1}{15} = \frac{1}{5} .

Question 134. Two events A and B have the probabilities 0.25 and 0.5 respectively. The probability that both A and B simultaneously is 0.12. then the probability that neither A and nor B occurs is ___.

0.13
0.38
0.63
0.37

Discuss Question

Answer: Option D. -> 0.37
:
D
Given:

P (A) = 0.25

P (B) = 0.5

P(A and B simultaneously happening) = 0.12

P (A \cup B) = P (A) + P (B) - P (A \cap B) P (A \cup B) = 0.25 + 0.5 - 0.12 = 0.63

P (¯ A \cap ¯ B) = P ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ (A \cup B) = 1 - P (A \cup B) = 1 - 0.63 = 0.37

Question 135. The probabilities of three mutually exclusive events A, B, C are given by

\frac{2}{3}, \frac{1}{4}, a n d \frac{1}{6}

respectively. This statement

Is true
is false
nothing can be said
could be either

Discuss Question

Answer: Option B. -> is false
:
B
Since the events A,B,C are mutually exclusive, we have

P (A ⋃ B ⋃ C) = \frac{2}{3} + \frac{1}{4} + \frac{1}{6} = \frac{13}{12} > 0

Which is not possible , hence the statement is false.

Question 136. A father has three children with at least one boy. The probability that he has two boys and one girl is

Discuss Question

Answer: Option B. -> 37
:
B
Let A be the event that father has at least one boy and B be the event that he has 2boys and one girl.

P (A) = P (1 b o y, 2 g i r l) + P (2 b o y, 1 g i r l) + P (3 b o y, n o g i r l) = \frac{3 C_{1} + 3 C_{2} + 3 C_{3}}{2^{3}} = \frac{7}{8}

P (A ⋂ B) = P (2 b o y, 1 g i r l) = \frac{3}{8}

H e n c e P (\frac{B}{A}) = \frac{P (A ⋂ B)}{P (A)} = \frac{3}{7} .

Question 137. From a bag containing 9 distinct white and 9 distinct black, 9 balls are drawn at random one by one, the drawn balls being replaced each time. The probability that at least four balls of each colour is in the draw, is

A little less than 12
a little greater than 12
12
13

Discuss Question

Answer: Option A. -> A little less than 12
:
A
Out of 9 distinct black and 9 distinct white balls, probability of drawing a white ball

= \frac{1}{2}

and drawing black ball is also

\frac{1}{2} .

for at least 4 of each colour in 9 draws with replacement, there are two cases,
Case(i)
P(getting 5 white, 4 black)

= 9 C_{5} {(\frac{1}{2})}^{9}

Cases (ii)
P(getting 4 white, 5 black)

= 9 C_{4} {(\frac{1}{2})}^{9}

These are exclusive so
P(atleast 4 of each colour)

= 9 C_{5} {(\frac{1}{2})}^{9} + 9 C_{4} {(\frac{1}{2})}^{9} = \frac{63}{128}

which is little less than

\frac{1}{2}

Question 138. Team A plays with 5 other teams exactly once. Assuming that for each match the probabilities of a win, draw and loss are equal, then

The probability that A wins and loses equal number of matches is 3481
The probability that A wins and loses equal number of matches is 1781
The probability that A wins more number of matches than it loses is 1781
The probability that A loses more number of matches than it wins is 1681

Discuss Question

Answer: Option B. -> The probability that A wins and loses equal number of matches is 1781
:
B
Probability of equal number of W and L is the probability of following
(0)W,(0)L +(1)W,(1)L+(2)W, (2)L

= {(\frac{1}{3})}^{5} +^{5} C_{1} .^{4} C_{1} {(\frac{1}{3})}^{5} +^{5} C_{2} .^{3} C_{2} {(\frac{1}{3})}^{5} = \frac{17}{81}

Question 139. A man is known to speak truth is 75% cases. If he throws an unbiased die and tells his friend that it is a six, then the probability that it is actually a six, is

Discuss Question

Answer: Option D. -> 38
:
D
Let E denote the event that a six occurs and A be the event theat the man reports that it is a six. We have