Question

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'll use P(A) • P(B) = P(A ∩ B)

Answer 2

@Vocaloid

Answer 3

@Hero I need help I don't know what to do next I will use P(A) • P(B) = P(A ∩ B) to start but what do I do next

Answer 4

@lowkey @Narad

Answer 5

P(A) • P(B) = P(A ∩ B)

Answer 6

its blank

Answer 7

oh

Answer 8

The answer to your question is 40/100 * 5/100
0.4 * 0.05
= 0.02
= 2%

Answer 9

I need to use P(A) • P(B) = P(A ∩ B)

Answer 10

help me use it, it should be easy because now we know the answer

Answer 11

Oh it's just p(0.4) times p(0.05)=P(0.4*0.05)

Answer 12

what does ∩ mean

Answer 13

You got sass.

Answer 14

p(0.4) times p(0.05)=P(0.4*0.05)

Answer 15

ya

Answer 16

p(0.4) times p(0.05)=P(0.4*0.05)

p(0.2)

Answer 17

is that whats next

Answer 18

p(0.4) times p(0.05)=P(0.4*0.05)

and then p(0.2) or do we do another step before we get 0.2

Answer 19

thats ok

Answer 20

thank you so much