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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