100+ TNPSC Statistics Exam Questions and Answers - 1

Ans: C

Here, n(S) = 52.

Let E = event of getting a queen of club or a king of heart.

Then, n(E) = 2.

∴ P(E) = $${n(E)} / {n(S)}$$ = $${2}/{52}$$ = $${1}/{26}$$.

Ans: D

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

Let E be the event of drawing 5 balls out of 9 non blue balls.

= ⁹C₅ = ⁹C_(9-5)

= ⁹C₄ = $${9× 8 × 7 × 6} / {4× 3 × 2 × 1}$$ = 126.

And, n(S) = ¹⁵C₅ = $${15 × 14 × 13 × 12 × 11} / {5 × 4 × 3 × 2 × 1}$$ = 3003.

∴ P(E) = $${n(E)} / {n(S)}$$ = $${126} / {3003}$$ = $${6}/{143}$$.

∴ Required probability = $$(1 - {6}/{143})$$ = $${137}/{143}$$.

Ans: B

Here n(S) = 6 × 6 = 36.

Let E = Event of getting a total more than 7.

= {(2, 6) (3, 5) (3, 6), (4, 4) (4, 5) (4, 6) (5, 3) (5, 6) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)}

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

Ans: C

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

Let E = event of getting a total of 10 or 11 = {(4, 6), (5, 5), (6, 4), (5, 6), (6, 5)}

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

Ans: C

Number of blue balls = 3balls

Number of green balls = 2 balls

Number of red balls = 5 balls

Total balls in the bag = 3 + 2 + 5 = 10

Total possible outcomes = Selection of 4 balls out of 10

balls = ¹⁰C₄ = $${10!}/{4 × (10 – 4)!}$$ = $${10 × 9 × 8 × 7} / {1 × 2 × 3 × 4}$$ = 210

Favorable outcomes = (Selection of 2 green balls out of 2 balls) × (selection of 2 balls out of 3 blue balls)

= ²C₂ × ³C₂

= 1 × 3 = 3

∴ Required probability = $${Favorable outcomes} / {Total possible outcomes}$$ = $${3}/{210}$$ = $${1}/{70}$$.