Question

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

1. both balls are red.
2. first ball is black and second is red.
3. one of them is black and other is red.

Answer 1

Let R = the event of drawing a red ball; B = the event of drawing a black ball

i. Probability of getting a red ball in the first draw P(R) = `8/(10 + 8) = 8/18 = 4/9`

Because the ball is put back again.

∴ The probability of getting a red ball in the second draw P(R) = `4/9`

∴ The probability of both balls being red = P(R). P(R) = `4/9 xx 4/9 = 16/81`

ii. Probability of getting a black ball in the first draw P(B) = `10/18 = 5/9`

The probability of getting a red ball in the second draw P(R) = `4/9`

∴ P(first black and second red) = P(B) . P(R) = `5/9 xx 4/9 = 20/81`

iii. P(one black and one red) = P(first black and second red) + P(first red and second black)

= `5/9 . 4/9 + 4/9 . 5/9`

= `40/81`