Question

A basketball player has made 60​% of his foul shots during the season. Assuming the shots are​ independent, find the probability that in​ tonight's game he does the following. ​a) Missed for the first time on his fourth attempt ​b) Makes his first basket on his fourth shot ​c) Makes his first basket on one of his first 3 shots

Answer 1

Answer:

a) 0.0864

b) 0.0384

c) 0.936

Explanation:

Probability that he makes his shot, P(A) = 0.6

Probability that he doesn't make the shot, P(A') = 1 - P(A) = 1 - 0.6 = 0.4

a) Probability that he Misses for the first time on his fourth attempt

P(A) × P(A) × P(A) × P(A') = 0.6 × 0.6 × 0.6 × 0.4 = 0.0864

b) Probability that he Makes his first basket on his fourth shot

P(A') × P(A') × P(A') × P(A) = 0.4 × 0.4 × 0.4 × 0.6 = 0.0384

c) Probability that he Makes his first basket on one of his first 3 shots

Sum of the probabilities that he makes all three first shots, two of the first three shots and one of the first three shots with the order irrelevant.

- Probability that he makes all first three shots = P(A) × P(A) × P(A) = 0.6 × 0.6 × 0.6 = 0.216

- Probability that he makes two out of the first three shots = (P(A) × P(A) × P(A')) + (P(A) × P(A') × P(A)) + (P(A') × P(A) × P(A)) = 3(0.6 × 0.6 × 0.4) = 0.432 (it's multipled by 3 because the probability is the same regardless of order)

- Probability of making only one of the first three shots = (P(A) × P(A') × P(A')) + (P(A') × P(A) × P(A')) + (P(A') × P(A') × P(A)) = 3(0.6 × 0.4 × 0.4) = 0.288 (It's multiplied by 3 too because the probabilities are the same too, regardless of order).

Probability that he Makes his first basket on one of his first 3 shots = 0.216 + 0.432 + 0.288 = 0.936