Question

PROBABILITY AND STATISTICS (WILL AWARD A MEDAL AND FAN )



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?

Answer 1

I need to be walked through the steps, I'm not very good at word problems.

Answer 2

There are two events (foul shots), A and B. The probability of A then B is\[P (A\text{ then } B) = P(A) \times P(B)\]Don't forget, the probabilities need to be expressed as decimals before doing any calculations with them.

Answer 3

So using that equation I would get my answer?

Answer 4

@ospreytriple

Answer 5

Yes. Show me what you get.

Answer 6

I'm not sure what to plug in to what

Answer 7

Event A is missing the first foul shot. What's P(A)?

Answer 8

40?

Answer 9

Express as a decimal.

Answer 10

.40

Answer 11

?

Answer 12

Right. Event B is missing the second foul shot. What's P(B) expressed as a decimal?

Answer 13

.05

Answer 14

Right. So the event of Event A happening then Event B happening is P(A) x P(B). What do you get. Express your final answer as a percent.

Answer 15

I'm not sure if i'm right, but 2 percent?

Answer 16

Correct. Well done! It's a small percentage because, even if the first shot is missed, the second shot is almost never missed. Make sense?

Answer 17

Yes! Thank you. Do you mind helping me with another one? If not that'll be fine.

Answer 18

OK. Delete this question and post your other one.