Abasketball player gets 2 free-throw shots when she is fouled by a player on the opposing team. she misses the first shot 40% of the time. when she misses the first shot, she misses the second shot 5% of the time. what is the probability of missing both free-throw shots?

The probability that the player misses the first shot =  40% = 0.4

When first shot is missed, the probability that she misses the second shot = 5% = 0.05

The probability of missing both shots will be the product of the probabilities of missing individual shots which will be:

Probability of missing both shots = 40% x 5% = 0.4 x 0.05 = 0.02 = 2%

Thus there is a 2% probability that the basketball player misses both free-throw shots.

Ithink the answer should be b.