Question

You choose a tile at random from a bag containing 4 A’s, 3 B’s, and 5 C’s. You replace the first tile in the bad and then choose again. Find P(B and B)

Answer 1

P(B and B) = 1/9

Explanation:

There are 4+3+5 = 12 tiles in total

The probability of selecting a B would be 4/12=1/3

When you are replacing a B tile, planning to pick a B tile again, the total tile count doesn't change. Therefore, because the two events are independent, their probabilities are multiplied and so P(B and B) = P(B) * P(B) = 1/3 * 1/3 = 1/9