Probability Aptitude Mcq Model Questions & Answers, Practice Test for ssc chsl tier 1 2024

Question :1

Two dice are thrown simultaneously. What is the probability of obtaining a multiple of 2 on one of them and a multiple of 3 on the other

Answer: (a)

Favourable cases for one are there i.e., 2, 4 and 6 and for other are two i.e., 3, 6.

Hence required probability = $[({3 × 2}/36) 2 - 1/36] = 11/36$

[As same way happen when dice changes numbers among themselves]

Question :2

Six dice are thrown. The probability that different number will turn up is

Answer: (b)

The number of ways of getting the different number

1, 2, ....., 6 in six dice = 6 !.

Total number of ways = $6^6$

Hence, required probability = ${6!}/6^6$

=${1 × 2 × 3 × 4 × 5 ×6}/6^6 = 5/324$

Question :3

A committee of 5 Students is to be chosen from 6 boys and 4 girls. Find the probability that the committee contains exactly 2 girls.

Answer: (c)

5 Students can be selected from 10 in $^10C_5$ ways.

∴ n (S) = $^10C_5 = {10!}/{5!.5!} = {10 × 9 × 8 × 7 × 6}/{5 × 4 × 3 × 2}$ = 252

Let A be the event that the committee includes exactly 2 girls and 3 boys.

The two girls. can be selected in $^4C_2$ ways and the 3 boys can be selected in $^6C_3$ ways.

∴ n(A) = $^4C_2 × ^6C_3$ = 6 × 20 = 120

∴ P(A) = ${n(A)}/{n(S)} = 120/252 = 10/21$

Question :4

In tossing three coins at a time, what is the probability of getting at most one head?

Answer: (b)

Possible samples are as follows

{HHH, HTH, HHT, THH, TTH, THT, HTT, TTT}

Let A be the event of getting one head.

Let B be the event of getting no head.

Favourable outcome for

A = {TTH, THT, HTT}

Favourable outcome for

B = {TTT}

Total no. of outcomes = 8

∴ P(A) = $3/8$, P(B) = $1/8$

∴ Required probability = Probability of getting one head + Probability of getting no head

= P(A) + P(B) = $3/8 + 1/8 = 4/8 = 1/2$

Question :5

The probability that a man can hit a target is 3/4. He tries 5 times. The probability that he will hit the target at least three times is

Answer: (d)

Required probability

= $^5C_3 (3/4)^3 (1/4)^2 + ^5C_4 (3/4)^4 (1/4) + ^5C_5 (3/4)^5 = 459/512$

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