Probability(Quantitative Aptitude ) Questions and answers for exam Preparation

Quantitative Aptitude

PROBABILITY MCQs

Probability, Probability I

Total Questions : 775 | Page 10 of 78 pages

Question 91. 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 92. One hundred identical coins, each with probability p of showing up heads are tossed once. If 0 < p < 1 and the probability of heads showing 50 coins is equal to that head showing 51 coins, then the value of p is

12
49101
50101
51101

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Answer: Option D. -> 51101
:
D
Let P1 and P2 be the respectively be the probability of heads showing 50 coins and head showing 51 coins

P_{1} = 100 C_{50} p^{50} (1 - p)^{50} a n d P_{2} = 100 C_{51} p^{51} (1 - p)^{49}

G i v e n P_{1} = P_{2}

\Rightarrow 100 C_{50} p^{50} (1 - p)^{50} = 100 C_{51} p^{51} (1 - p)^{49}

\Rightarrow \frac{1 - p}{p} = \frac{50}{51} \Rightarrow p = \frac{51}{101} .

Question 93. If

\frac{(1 + 4 p)}{4}, \frac{(1 - p)}{2}, and \frac{(1 - 2 p)}{2}

are the probabilities of three mutually exclusive events , then the value of p is

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Answer: Option A. -> 12
:
A

\frac{1 + 4 p}{4}, \frac{1 - p}{2}, \frac{1 - 2 p}{2}

are the probabilities of three mutually exclusive events.

∴ 0 \leq \frac{1 + 4 p}{p} \leq 1, 0 \leq \frac{1 - p}{2} \leq 1

0 \leq \frac{1 - 2 p}{2} \leq 1 a n d 0 \leq (\frac{1 + 4 p}{4}) + (\frac{1 - p}{2}) + \frac{1 - 2 p}{2} \leq 1

∴ \frac{- 1}{4} \leq p \leq \frac{3}{4} 1 - \leq p \leq 1, \frac{1}{2} \leq p \leq \frac{1}{2} a n d p \leq \frac{5}{2}

\Rightarrow p = \frac{1}{2}

Question 94. 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

P (A ⋃ B) = P (A) + P (B) - P (A ⋂ B) = 0.25 + 0.5 - 0.12 = 0.63

Therefore required probability = 1 - P(A intersection B) = 0.37

Question 95. Seven coupons are selected at random one at a time with replacement from 15 coupons numbered 1 to 15. The probability that the largest number appearing on a selected coupon is 9, is

(916)6
(815)7
(35)7
(35)7 - (815)7

Discuss Question

Answer: Option D. -> (35)7 - (815)7
:
D
Each coupon can be selected in 15 ways. The total number of ways of choosing 7 copouns is

15^{7}

. If largest number is 9, then the selected numbers have to be from 1 to 9 excluding those consisting of only 1 to 8.
Probability desired is

\frac{9^{7} - 8^{7}}{15^{7}}

= {(\frac{3}{5})}^{7} - {(\frac{8}{15})}^{7}

Question 96. 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 97. A random variable X has the following probability distribution

\begin{matrix} X & P (X = x) & X & P (X = x) 0 & λ & 5 & 11 λ 1 & 3 λ & 6 & 13 λ 2 & 5 λ & 7 & 15 λ 3 & 7 λ & 8 & 17 λ 4 & 9 λ \end{matrix}

then,

λ

is equal to

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Answer: Option A. -> 181
:
A

\sum_{x = 0}^{8} P (X) = 1 \Rightarrow P (0) + P (1) + P (2) + P (3) + P (4) + P (5) + P (6) + P (7) + P (8) = 1 = λ + 3 λ + 5 λ + 7 λ + 9 λ + 11 λ + 13 λ + 15 λ + 17 λ = 1 \Rightarrow \frac{9}{2} (λ + 17 λ) = 1 \Rightarrow 81 λ = 1 ∴ λ = \frac{1}{81}

Question 98. A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to that getting 9 heads, then the probability of getting 3 heads is

35212
35214
7212
7214

Discuss Question

Answer: Option A. -> 35212
:
A
Probability of r successes in n trials is equal to

^{n} C_{r} p^{r} q^{n - r}

, where p is the probability of success in one trial and q is the probability of failure in one trial.
Let the coin be tossed n times. Here p and q are both 1/2

P (g e t t i n g 7 h e a d s) = n C_{7} {(\frac{1}{2})}^{n} a n d P (g e t t i n g 9 h e a d s) = n C_{9} {(\frac{1}{2})}^{n} G i v e n, P (7 h e a d s) = P (9 h e a d s) \Rightarrow n C_{7} = n C_{9} \Rightarrow n = 16 ∴ P (3 h e a d s) = 16 C_{3} {(\frac{1}{2})}^{16} = \frac{35}{2^{12}} .

Question 99. A five digit number is formed with digits 0. 1. 2. 3. 4 without repetition. A number is selected at random, then the probability that it is divisible by 4 is

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Answer: Option B. -> 516
:
B
The number formed is divisible by 4 if the last two digits are 04,40,34,32,20,12.
Therefore total number of favorable ways = 3! + 3! +4 + 4+ 3! + 4 = 30 (This is the sum of number of ways in which the first two digits can be formed)
Total numbers that can be formed = 5! – 4! = 96 ( Number of ways of arranging 5 digits - Number of ways in which zero comes as the first digit)
Therefore required probability