# 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}\$\$.

Related Questions