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

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

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

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

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

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