Question

Suppose you have an urn containing 14 BLUE balls and 10 GREEN balls. You pick 6 balls out of the urn.

If the BLUE balls are numbered 1 to 14 and the GREEN balls, 1 to 10, what is probability that you can pick 3 odd BLUE balls and 2 even GREEN balls?

Is the solution 1/2 x C(14,3) x 1/2 C(10,2) divided by C(24,5)?

Answer 1

Total number of balls : 14+ 10 =24

Number of ways of selecting 6 balls out of 24 is C(24,6) = 134596

Out of 14 blue balls, 7 are odd and out of 10 green balls 5 are even.

Number of ways of selecting 3 odd balls out of 7 and 2 green balls out of 5 even green balls is C(7,3)*C(5,2).

Since 7 blue balls are even and 5 green balls are odd so selecting 1 balls out of 7+5 =12 balls is C(12,1).

Therefore number of ways of selecting 3 odd BLUE balls and 2 even GREEN balls is

C(7,3)*C(5,2)*C(12,1) = 4200

The probability that you can pick 3 odd BLUE balls and 2 even GREEN balls is

[C(7,3)*C(5,2)*C(12,1)] / C(24,6) = 4200 / 134596 = 0.0312