6.11 In basketball, sometimes a player shoots free throws \"one and one,\" meani

Question

6.11 In basketball, sometimes a player shoots free throws "one and one," meaning if the first free throw goes in the player shoots a second, but if the first misses the player does not shoot a second. Sometimes the player shoots two free throws (the player shoots the second whether or not the first one went in). Each successful free throw is worth one point. Assume the free throws are independent, each with probability p of going in. a. What is the expected number of points shooting "one and one"? b. What is the expected number of points shooting two free throws? c. What is the value of p that maximizes the difference between the two answers above?

Explanation / Answer

(a.) Let X be the points taken through one and one

for X=1, P(X=1) = p*(1-p) {first gets in and second does not}

for X=2, P(X=2) = p*p

Thus E(X) = 1*p*(1-p) + 2*p*p = p- p*p + 2*p*p = p(1+p)

(b)

for X=1, P(X=1) = p*(1-p) + (1-p)*p = 2*p*(1-p) {first gets in and second does not or second goes in and first does not}

for X=2, P(X=2) = p*p

Thus E(X) = 1*2*p*(1-p) + 2*p*p = 2p- 2*p*p + 2*p*p = 2p

(c) Difference = p^2 -p

it maximises where d(difference)/dp =0 {differentiation}

2*p-1 = 0

thus, p =0.5

X=0 is ignored because it will not contribute to the expected number of points.

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