100+ Probability TNPSC Previous Year Questions - 1

Ans: A

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}.

Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}.

∴ P (E) = $${n(E)}/{n(S)}$$ = $${4}/{8}$$ = $${1}/{2}$$.

Ans: B

In a simultaneously throw of two dice, we have n(S) = (6 × 6) = 36.

Let E = event of getting two numbers whose product is even.

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

∴ n(E) = 27.

∴ P(E) = $${n(E)}/ {n(S)}$$ = $${27}/{36}$$ = $${3}/{4}$$.

Ans: C

Total number of balls = (8 + 7 + 6) = 21.

Let E = Event that the ball drawn is neither red nor green

= Event that the ball drawn is red.

∴ n(E) = 8.

∴ P(E) = $${8}/{21}$$.

Ans: C

Total number of balls = (2 + 3 + 2) = 7.

Let E be the event of drawing 2 non blue balls.

Then, n(E) = ⁵C₂ = $${5 × 4} / {2 × 1}$$ = 10.

Also, n(S) = ⁷C₂ = $${7 × 6} / {2 × 1}$$ = 21.

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

Ans: C

Clearly, n(S) = (6 × 6) = 36.

Let E be the event that the sum is a prime number.

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

∴ n(E) = 15.

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