Answer:
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)
<u>Step 1. </u>
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
<u>Step 2.</u>
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!
