100+ Probability Questions and Answers for SSC Exams - 1

Ans: D

Here S = {HH, HT, TH, TT}.

Let E = event of getting at most one head.

∴ E = {TT, HT, TH}

∴ P(E) = $${n(E)} / {n(S)}$$ = $${3}/{4}$$.

Ans: C

Here, n(S) = 52.

There are 26 red cards (including 2 kings) and there are 2 more kings.

Let E = event of getting a red card or a king.

Then, n(E) = 28.

∴ P(E) = $${n(E)} / {n(S)}$$ = $${28}/{52}$$ = $${7}/{13}$$.

Ans: B

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

Let E be the event of drawing 4 balls such that 2 are red and 2 are green.

Then, n(E) = (²C₂ × ⁴C₂) = $$(1 × {4 × 3}/{2 × 1})$$ = 6.

And n(S) = ¹⁵C₄ = $${15 × 14 × 13 × 12}/ {4 × 3 × 2 × 1}$$ = 1365

∴ P(E) = $${n(E)} / {n(S)}$$ = $${6}/ {1365}$$ = $${2}/{455}$$.

Ans: C

Let S be the sample space and let E be the event of selecting 2 boys and 1 girl.

Then, n(S) = number of ways of selecting 3 students

out of 25 = ²⁵C₃ = $${25 × 24 × 23} / {3 × 2 × 1}$$ = 2300.

And, n(E) = (¹⁵C₂ × ¹⁰C₁) = $$({15 × 14} / {2 × 1} × 10)$$ = 1050.

∴ P(E) = $${n(E)}/{n(S)}$$ = $${1050}/{2300}$$ = $${21}/{46}$$.

Ans: B

Number of Red balls = 4

Number of Blue balls = 6

Number of Pink balls = 8

Total number of balls = 4 + 6 + 8 = 18.

Required probability = $${4/18} × {11/20} + {6/18} × {11/20}$$

                             = $${11}/{20}[({4/8} + {6/18})]$$

                             = $${11}/{20} × {10}/{18}$$ = $${11}/{36}$$.