7000+ Hibret Bank Aptitude Questions and Answers Pdf - 1

Ans: A

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

Number of exhaustive cases = 4

There are two favourable cases HT, TH.

∴ P (exactly 1 tail) = $${2}/{4}$$ = $${1}/{2}$$.

Ans: C

Here, n(S) = 6 x 6 = 36.

Let E = Event of getting a total more than 7

= {(2, 6), (3, 5), (3, 6), (4,4), (4, 5), (4, 6), (5, 3), (5, 4),
(5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

∴ P(E) =$${ n(E)} / {n(S)}$$ = $${15}/ {36}$$ = $${5}/{12}$$.

Ans: D

Total number of marbles = (6 + 4 + 2 + 3) = 15.

Let E be the event of drawing 2 blue and 1 yellow marble.

Then, n(E) = (⁴C₂ × ³C₁) = $${4 × 3} / {2 × 1} × 3$$ = 18.

Also, n(S) = ¹⁵C₃ = $${15 × 14 × 13} / {3 × 2 × 1}$$ = 455.

∴ P(E) = $${n(E)} / {n(S)}$$ = $${18}/{455}$$.

Ans: C

Number of ways of selecting 4 students out of 15

students = ¹⁵C₄ = $${15 × 14 × 13 × 12} / {1 × 2 × 3 × 4}$$ = 1365

The number of ways of selecting 4 students in which no student belongs to Karnataka = ¹⁰C₄

∴ No of ways of selecting at least 1 student from Karnataka = ¹⁵C₄ - ¹⁰C₄ = 1155

∴ Required probability = $${1155}/{1365}$$ = $${77}/{91}$$ = $${11}/{13}$$

Ans: B

n(s) = Total possible outcomes

= ¹²C₂ = $${12 × 11} / {1 × 2}$$ = 66

n(E) = Favourable number of cases

= ⁴C₂ = $${4 × 3} / {1 × 2}$$ = 6

Required probability = $${n(E)} / {n(S)}$$ = $${6}/{66}$$ = $${1}/{11}$$.