7000+ Probability Aptitude Questions with Solutions Pdf - 1

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 = $${5! × 3!} /{7!}$$ = $${3 × 2 × 1}/ {7 × 6}$$ = $${1}/{7}$$.

Ans: B

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

Number of exhaustive cases = 4.

There is only one favourable case HH.

∴ P (no tails) = $${1}/{4}$$.

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 = $${6}/ {36}$$ = $${1}/{6}$$.

Ans: C

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

= 0.38 + 0.30 = 0.68.

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) = $${4}/{216}$$ = $${1}/{54}$$