An urn initially contains 5 white and 7 black balls Each tim

An urn initially contains 5 white and 7 black balls. Each time a ball is selected, its color is noted and it is replaced in the urn along with 2 other balls of the same color. Compute the probability that (a) the first 2 balls selected are black and the next 2 are white; (b) Of the first 4 balls selected, exactly 2 are black.

Solution

Total initial urns = 12

Let BBWW denote first 2 balls selected are black and the next 2 are white

a. P(BBWW) = (7/12)*(9/14)*(5/16)*(7/18) = 35/768 = 0.0456

b. P (exactly 2 black in 1st 4 balls) has possible outcomes:

P(BBWW) + P(BWBW) + P(BWWB) + P(WBBW) + P(WBWB) + P(WWBB)

= (7/12)*(9/14)*(5/16)*(7/18) + (7/12)*(5/14)*(9/16)*(7/18) + (7/12)*(5/14)*(7/16)*(9/18) + (5/12)*(7/14)*(9/16)*(7/18)

+ (5/12)*(7/14)*(7/16)*(9/18) + (5/12)*(7/14)*(7/16)*(9/18) = 35/128 = 0.2734