A pro basketball player is a poorâ free-throw shooter. Consider situations in which he shoots a pair of free throws. The probability that he makes the first free throw is 0.48. Given that he makes theâ first, suppose the probability that he makes the second is 0.62. Given that he misses theâ first, suppose the probability that he makes the second one is 0.38.

1. Find the probability that he makes one of the two free throws using the multiplicative rule with the two possible ways he can do this.

The probability that he makes one of the two free throws is 0.38

Step-by-step explanation:

Hello!

Considering the situation:

A pro basketball player shoots two free throws.

The following events are determined:

A: "He makes the first free throw"

Ac: "He doesn't make the first free throw"

B: "He makes the second free throw"

Bc: "He doesn't make the second free throw"

It is known that

P(A)= 0.48

P(B/A)= 0.62

P(B/Ac)= 0.38

You need to calculate the probability that he makes one of the two free throws.

There are two possibilities, that "he makes the first throw but fails the second" or that "he fails the first throw and makes the second"

Symbolically:

P(A∩Bc) + P(Ac∩B)

Step 1.

P(A)= 0.48

P(Ac)= 1 - P(A)= 1 - 0.48= 0.52

P(Ac∩B) = P(Ac) * P(B/Ac)= 0.52*0.38= 0.1976≅ 0.20

Step 2.

P(A∩B)= P(A)*P(B/A)= 0.48*0.62= 0.2976≅ 0.30

P(A)= P(A∩B) + P(A∩Bc)

P(A∩Bc)= P(A) - P(A∩B)= 0.48 - 0.30= 0.18

Step 3

P(Ac∩B) + P(A∩Bc) = 0.20 + 0.18= 0.38

I hope this helps!

Ok You're probably here from K12 so heres the

Conclusion: If a polygon undergoes a rigid transformation, then the image and pre-image have equal perimeters.

I just took the test and got this right, I'm a K12 student as well.

Hope this helps :)