# TNPSC Group 1, 2, 2a, 4, VAO Maths Question Paper with Answers - 1

Question: 1

Two cards are drawn from a pack of 52 cards. The probability that one is a spade and one is a heart, is

(A) \$\${3}/{20}\$\$

(B) \$\${29}/{34}\$\$

(C) \$\${47}/{100}\$\$

(D) \$\${13}/{102}\$\$

Ans: D

Let S be the sample space.

Then, n(S) = 52C2 = \$\$({52 × 51}/{2× 1})\$\$ = 1326.

Let E = event of getting 1 spade and 1 heart.

n(E) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13

= (13C1 × 13C1) = 13 × 13 = 169.

∴ P(E) = \$\${n(E)} / {n(S)}\$\$ = \$\${169}/{1326}\$\$ = \$\${13}/{102}\$\$.

Question: 2

An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If two marbles are drawn at random from the urn, what is the probability that both are red?

(A) \$\${1}/{5}\$\$

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

(C) \$\${1}/{7}\$\$

(D) \$\${2}/{5}\$\$

Ans: C

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

Let E be the event of drawing 2 red balls.

Then, n(E) = 6C2 = \$\${6 × 5} / {2 × 1}\$\$ = 15.

Also, n(S) = 15C2 = \$\${15 × 14} / {2 × 1}\$\$ = 105.

∴ P(E) = \$\${n(E)} / {n(S)}\$\$ = \$\${15}/{105}\$\$ = \$\${1}/{7}\$\$.

Question: 3

An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If two marbles are picked up at random, what is the probability that either both are green or both are yellow?

(A) \$\${5}/{91}\$\$

(B) \$\${1}/{35}\$\$

(C) \$\${1}/{3}\$\$

(D) \$\${4}/{105}\$\$

Ans: D

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

Let E be the event of drawing 2 marbles such that either both are green or both are yellow.

Then, n(E) = (2C1 + 3C2) = (1 + 3C1) = (1 + 3) = 4. And,

n(S) = 15C2 = \$\${15 × 14} / {2 × 1}\$\$ = 105.

∴ P(E) = \$\${n(E)}/{n(S)}\$\$ = \$\${4}/{105}\$\$.

Question: 4

A box contains 10 black and 10 white balls. What is the probability of drawing 2 balls of the same colour?

(A) \$\${9}/{19}\$\$

(B) \$\${9}/{38}\$\$

(C) \$\${10}/{19}\$\$

(D) \$\${5}/{19}\$\$

Ans: A

Total number of balls = (10 + 10) = 20.

Let E be the event of drawing 2 balls of the same colour.

n(E) = number of ways of drawing 2 black balls or 2 white balls.

n(E) = (10C2 + 10C2) = 2 × 10C2 = \$\$2 × {10 × 9}/ {2 × 1}\$\$ = 90.

n(S) = number of ways of drawing 2 balls out of 20

= 20C2 = \$\${20 × 19}/ {2 × 1}\$\$ = 190.

P(E) = \$\${n(E)} / {n(S)}\$\$ = \$\${90}/{190}\$\$ = \$\${9}/{19}\$\$.

Question: 5

A basket contains 4 red, 5 blue and 3 green marbles. If 2 marbles are drawn at random from the basket, what is the probability that both are red?

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

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

(C) \$\${1}/{11}\$\$

(D) \$\${3}/{7}\$\$

Ans: C

Total number of balls = (4 + 5 + 3) = 12.

Let E be the event of drawing 2 red balls.

Then, n(E) = 4C2 = \$\${4 × 3} / {2 × 1}\$\$ = 6.

Also, n(S) = 12C2 = \$\${12 × 11} / {2 × 1}\$\$ = 66.

∴ P(E) = \$\${n(E)} / {n(S)}\$\$ = \$\${6}/{66}\$\$ = \$\${1}/{11}\$\$.

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