## anonymous one year ago A basketball 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?

1. anonymous

?

2. misty1212

multiply

3. kropot72

Let A be the event 'misses the first shot' and let B be the event 'misses the second shot. P(A) = 0.4 P(B|A) = 0.05 $\large P(A \cap B)=P(A) \times P(B|A)$

4. anonymous

.02 ?

5. kropot72

Correct.

6. anonymous

7. kropot72

Yes, that value is the probability of missing both free-throw shots.

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