7000+ Probability Questions for IBPS Clerk Pdf - 1

Ans: B

Clearly, n(s) = 6 × 6 = 36.

Let E be the event that the sum of the numbers on the two faces is divisible by 4 or 6.

Then E = {(1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5),(4, 2), (4, 4), (5, 1), (5, 3), (6, 2), ((6, 6)}

∴ n(E) = 14.

Hence, P(E) = $${n(E)} / {n(S)}$$ = $${14}/ {36}$$ = $${7}/ {18}$$.

Ans: B

Total number of ways of selecting 4 children out of 8

= ⁸C₄ = $${8 × 7 × 6 × 5} / {1 × 2 × 3 × 4}$$ = 70

Number of ways of selecting 4 girls out of 5 = ⁵C₄ = 5

Required probability = $${5}/{70}$$ = $${1}/{14}$$.

Ans: A

There are 4 favourable cases TTTH, TTHT, THTT, HTTT.

∴ P(exactly 3 tails) = $${4}/{16}$$ =$${1}/{4}$$.

Ans: A

The selection of 1 boy and 3 girls can be done in ⁶C₁ × ⁴C₃ = 6 × 4 = 24 ways

Selection of 4 girls and no boys can be done in ⁶C₀ × ⁴C₄ = 1 × 1 = 1 way.

∴ n(E) = 24 + 1 = 25 ways.

Without any restriction the group can be formed in ¹⁰C₄ = $${10 × 9 × 8 × 7} / {1 × 2 × 3 × 4}$$ = 210 ways.

∴ P(E) = $${n(E)} / {n(S)}$$ =$${25} / {210}$$ = $${5}/{42}$$.

(i) Probability that both balls drawn were red = $${5/12} × {5/12}$$ = $${25}/{144}$$

(ii) Probability that both balls drawn were green = $${7/12} × {7/12}$$ = $${49}/ {144}$$

(iii) Probability that the first ball drawn was red and second green $${5/12} ×{7/12}$$ = $${35}/{144}$$

(iv) Probability that the first ball drawn was green and second red $${7/12} × {5/12}$$ = $${35}/{144}$$