Question

Suppose a ball is drawn at random from a box containing three white and three black
balls. After a ball is drawn, it is then replaced and another drawn. What is the
probability that of the first four balls drawn, exactly two are white?

Answer 1

$$P(WWBB)+P(WBBW)+P(WBWB)+$$

$$+P(BWWB)+P(BWBW)+P(BBWW)=$$

$$=({3 \over 6})({3\over 6})({3 \over 6})({3 \over 6})+({3 \over 6})({3\over 6})({3 \over 6})({3 \over 6})+$$

$$+({3 \over 6})({3\over 6})({3 \over 6})({3 \over 6})+({3 \over 6})({3\over 6})({3 \over 6})({3 \over 6})+$$

$$+({3 \over 6})({3\over 6})({3 \over 6})({3 \over 6})+({3 \over 6})({3\over 6})({3 \over 6})({3 \over 6})=$$

$$=6({1\over 16})={3\over 8}=0.375$$