# 7000+ Probability Aptitude Questions with Solutions Pdf - 1

Question: 1

The letters of word 'SOCIETY' are placed in a row. What is the probability that three come together?

(A) \$\${1}/{7}\$\$

(B) \$\${2}/{7}\$\$

(C) \$\${3}/{7}\$\$

(D) \$\${4}/{7}\$\$

Ans: A

There are 7 letters in the word ‘SOCIETY’ which can be arranged in 71 ways. Considering the three vowels in the word ‘SOCIETY’ as one letter, we can arrange 5 letters in a row in 5! ways. Also, three vowels can themselves be arranged in 3! ways.

∴ The total number of arrangements in which three vowels come together are 5! × 3!

Hence, the required probability = \$\${5! × 3!} /{7!}\$\$ = \$\${3 × 2 × 1}/ {7 × 6}\$\$ = \$\${1}/{7}\$\$.

Question: 2

In a simultaneous toss of 2 coins, then find the probability of no tail

(A) \$\${1}/{2}\$\$

(B) \$\${1}/{4}\$\$

(C) \$\${2}/{3}\$\$

(D) \$\${3}/{4}\$\$

Ans: B

Sample space S = {HH, HT, TH, TT}

Number of exhaustive cases = 4.

There is only one favourable case HH.

∴ P (no tails) = \$\${1}/{4}\$\$.

Question: 3

In a single throw of 2 dice, the probability of getting a total of 7.

(A) \$\${1}/{6}\$\$

(B) \$\${1}/{18}\$\$

(C) \$\${5}/{6}\$\$

(D) \$\${5}/{36}\$\$

Ans: A

A total of 7 may be obtained in 6 ways, viz,
(1, 6) (2, 5), (3, 4), (5, 2), (6, 1).

∴ Required probability = \$\${6}/ {36}\$\$ = \$\${1}/{6}\$\$.

Question: 4

The probabilities that a student will receive an A, B, C or D grade are 0.30, 0.38, 0.22 and 0.01 respectively. What is the probability that the student will receive at least B grade?

(A) 0.38

(B) 0.42

(C) 0.68

(D) None of these

Ans: C

P(at least B grade) = P(B grade) + P(Agrade)

= 0.38 + 0.30 = 0.68.

Question: 5

In a single throw of 3 dice, find the probability of getting a total of 17 or 18.

(A) \$\${1}/{18}\$\$

(B) \$\${1}/{4}\$\$

(C) \$\${1}/{36}\$\$

(D) \$\${1}/{54}\$\$

Ans: D

Number of exhaustive cases in a single throw of three dice

= 6 × 6 × 6 = 216

Cases favourable to a total of 17 are (5, 6, 6), (6, 5, 6), (6, 6, 5)

Number of cases favourbale to a total of 17 or 18 is 4.

∴ P( a total of 17 or 18) = \$\${4}/{216}\$\$ = \$\${1}/{54}\$\$

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