100+ Probability Aptitude Questions and Answers - 2

Ans: C

Number of red balls = 4.

Number of yellow ball = 5.

Number of pink ball = 6.

Total number of balls = 4 + 5 + 6 = 15.

Total possible outcomes = selection of 2 balls out of 15 balls

= ¹⁵C₂ = $${15!} / {2! (15 – 2)!}$$ = $${15!} / {2! × 13!}$$ = $${15 × 14} / {1 × 2}$$ = 105.

Total favourable outcomes = selection of 2 balls out of 4 orange and 6 pink balls.

¹⁰C₂ = $${10!} / {2! (10 – 2)!}$$ = $${10!} / {2! (8!)} $$ = $${10 × 9} / {1 × 2}$$ = 45.

∴ Required probability = $${45} / {105}$$ = $${3} / {7}$$.

Ans: D

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

Let E = event of getting at least one head = {HT, TH, HH}.

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

Ans: B

When a die is thrown, we have S = {1, 2, 3, 4, 5, 6}.

Let E = event of getting a number greater than 4 = {5, 6}.

∴ P(E) = $${n(E)} / {n(S)}$$ = $${2} / {6}$$ = $${1} / {3}$$.

Ans: B

Ans: D

In two throws of a die, n(S) = (6 × 6) = 36.

Let E = event of getting a sum 9 = {(3, 6), (4, 5), (5, 4) (6, 3)}.

∴ P(E) = $${n(E)} / {n(S)}$$ = $${4} / {36}$$ = $${1} /{9}$$.