100+ Probability TNPSC Previous Year Questions - 1

Question: 1

Three unbiased coins are tossed. What is the probability of getting at least 2 heads?

(A) $${1}/{2}$$

(B) $${1}/{3}$$

(C) $${1}/{4}$$

(D) $${1}/{8}$$

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

Question: 2

Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

(A) $${1}/{2}$$

(B) $${3}/{4}$$

(C) $${3}/{8}$$

(D) $${5}/{16}$$

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

Question: 3

In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

(A) $${2}/{3}$$

(B) $${3}/{4}$$

(C) $${8}/{21}$$

(D) $${9}/{23}$$

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) = $${8}/{21}$$.

Question: 4

An urn contains 3 red, 3 green and 2 blue balls. If 2 balls are drawn at random, find the probability that no ball is blue.

(A) $${2}/{7}$$

(B) $${5}/{7}$$

(C) $${10}/{21}$$

(D) $${11}/{21}$$

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

Question: 5

Two dice are tossed. The probability that the total score is a prime number is

(A) $${1}/{2}$$

(B) $${1}/{6}$$

(C) $${5}/{12}$$

(D) $${7}/{9}$$

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

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