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

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

seeing the graph would

d) physical means can be used to separate the components of air.