There are 2 red and 3 black balls in a bag. 3 balls are taken out at random from the bag. Find the probability of getting 2 red and 1 black ball or 1 red and 2 black balls.

#### Solution

There are 2 + 3 = 5 balls in the bag and 3 balls can be drawn out of these in ^{5}C_{3}= `(5 xx 4 xx 3)/(1 xx 2 xx 3)` = 10 ways.

∴ n(S) = 10

Let A be the event that 2 balls are red and 1 ball is black

2 red balls can be drawn out of 2 red balls in ^{2}C_{2} = 1 way and 1 black ball can be drawn out of 3 black balls in ^{3}C_{1} = 3 ways.

∴ n(A) = ^{2}C_{2} × ^{3}C_{1} = 1 × 3 = 3

∴ P(A) = `("n"("A"))/("n"("S")) = 3/10`

Let B be the event that 1 ball is red and 2 balls are black

1 red ball out of 2 red balls can be drawn in ^{2}C_{1} = 2 ways and 2 black balls out of 3 black balls can be drawn in ^{3}C_{2} = `(3 xx 2)/(1 xx 2)` = 3 ways.

∴ n(B) = ^{2}C_{1} × ^{3}C_{2} = 2 × 3 = 6

∴ P(B) = `("n"("B"))/("n"("S")) = 6/10`

Since A and B are mutually exclusive and exhaustive events

∴ P(A ∩ B) = 0

∴ Required probability = P(A ∪ B) = P(A) + P(B)

= `3/10 + 6/10`

= `9/10`