Question 1009654: 5. A basketball player has an overall 60% average of success while shooting free
throws.
(a) After a foul, she gets ready to make two free throws. One of your students
says “she’s bound to make at least one of the two, since 2×60% > 100%”.
What is the problem with this argument? Explain.
(b) If each successful free throw is worth one point, what is the expected
number of points she’ll make after the foul?
Please help me understand the reasoning behind the solution
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
Question:
5. A basketball player has an overall 60% average of success while shooting free
throws.
(a) After a foul, she gets ready to make two free throws. One of your students
says “she’s bound to make at least one of the two, since 2×60% > 100%”.
What is the problem with this argument? Explain.
(b) If each successful free throw is worth one point, what is the expected
number of points she’ll make after the foul?
Please help me understand the reasoning behind the solution
Solution:
A.
First, take note that two tries do not double the probability of success, since probabilities cannot be greater than one. In fact, the sum of probability of all possible outcomes add up to 1. Therefore, we must search for all possible outcomes and determine the probability of each event happening.
B.
We define S=successful throw, and F=unsuccessful throw.
The four possible outcomes are SS, SF, FS, FF.
The corresponding probabilities are:
P(SS)=0.6*0.6=0.36
P(SF)=0.6*0.4=0.24
P(FS)=0.4*0.6=0.24
P(FF)=0.4*0.4=0.16
We define N(x) as the number of points attributed to event x.
The expected number of points are therefore
E(x)=sum P(x)*N(x)
=0.36*2+0.24*1+0.24*1+0.16*0
=1.2
So the expected number of points is 1.2
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