NIB International Bank Exam Questions and Answers Pdf - 1

Ans: C

Total number of possible events n(s) = {(H, H), (H, T), (T, H), (T, T)} = 4

Event of getting at least one head E = {(H, H), (H, T), (T, H)} = 3

P(E) = $${3}/{4}$$

Ans: B

n(S) = 4 i.e. {(T, T), (H, H), (H, T), (T, H)}

P(occurrence of at least one head) = $${n(A)} / {n(S)}$$ = $${3}/{4}$$

P(occurrence of at least one tail) = $${n(A)} / {n(S)}$$ = $${3}/{4}$$

Ans: C

Total number of balls = 5 + 4 + 3 = 12

P(balls are not of the same colour) = ?

n(S) = ¹²C₃ = $${12 × 11 × 10} / {1 × 2 × 3}$$ = 220

i.e. 3 balls can be drawn from the given 12 balls in 220 ways.

If all are of same colour, it can be done in ⁵C₃ + ⁴C₃ + ³C₃ = 10 + 4 + 1 = 15.

P(all balls are of same colour) + P(all balls are not of same colour) = 1.

∴ P(all balls are not of same colour) = 1 – P(all balls are not of same colour)

= 1 - $${15} / {220}$$ = $${205}/ {220}$$ = $${41}/ {44}$$

Ans: B

Total number of possible events = 2 + 3 + 2 = 7

The number of ways in which 2 balls can be drawn out of given 7 = ⁷C₂ = $${7 × 6} / {1 × 2}$$ = 21

i.e. n(S) = 21

E = event of getting balls not to be blue

n(E) = ways of drawing 2 balls from given 5 = ⁵C₂ = $${5 × 4} / {1 × 2}$$ = 10

P(E) = $${n(E)} / {n(S)}$$ = $${10}/ {21}$$

Ans: D

Number of possible events to draw three balls from 4 white, 5 red and 6 blue balls

= ¹⁵C₃ = $${15 × 14 × 13} / {1 × 2 × 3}$$ = 5 × 7 × 13 = 455

E = Event of getting 3 balls red

The number of favourable events to get 3 red balls = ⁵C₃ = $${5 × 4} / {1 × 2}$$ = 10

∴ P(E) = $${n(E)}/{n(S)}$$ = $${10}/{455}$$ = $${2}/{91}$$