Question

The best player on a basketball team makes 80​% of all free throws. The​ second-best player makes 75​% of all free throws. The​ third-best player makes 70​% of all free throws. Based on their experimental​ probabilities, estimate the number of free throws each player will make in his or her next 60 attempts.

Answer 1

Expected number of free throws in 60 attempts:

Best player = 48

2nd best player = 45

3rd best player = 42

Explanation:

Solution:-

- The probability that best player makes free throw, p1 = 0.8

- The probability that second-best player makes free throw, p2 = 0.75

- The probability that third-best player makes free throw, p3 = 0.70

- Total number of attempts made in free throws, n = 60.

- The estimated number of free throws that any player makes is defined by:

E ( Xi ) = n*pi

Where, Xi = Player rank

pi = Player rank probability

- Expected value for best player making the free throws would be:

E (X1) = n*p1

= 60*0.8

= 48 free throws

- Expected value for second-best player making the free throws would be:

E (X2) = n*p2

= 60*0.75

= 45 free throws

- Expected value for third-best player making the free throws would be:

E (X3) = n*p3

= 60*0.70

= 42 free throws