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

To find the probability of different scenarios, we use the concept of independent events. For (a) and (b), we multiply the probability of missing the first three shots with the probability of making the fourth shot. For (c),we subtract the probability of missing on all three shots from 1.

Step-by-step explanation:

To find the probability of different scenarios in tonight's game, we need to use the concept of independent events. We are given that the basketball player has a 60% foul shot success rate. This means that the probability of him making each shot is 0.6, and the probability of him missing each shot is 0.4.

a) The probability of him missing for the first time on his fourth attempt can be calculated by multiplying the probability of him missing the first three shots (0.4 x 0.4 x 0.4) with the probability of him making the fourth shot (0.6):

P(X=3) = (0.4 x 0.4 x 0.4) x 0.6 = 0.0384

b) The probability of him making his first basket on his fourth shot can be calculated by multiplying the probability of him missing the first three shots (0.4 x 0.4 x 0.4) with the probability of him making the fourth shot (0.6):

P(X=3) = (0.4 x 0.4 x 0.4) x 0.6 = 0.0384

c) The probability of him making his first basket on one of his first three shots can be calculated by subtracting the probability of him missing on all three shots:

P(X<=3) = 1 - (0.4 x 0.4 x 0.4) = 1 - 0.064 = 0.936

Answer 2

a) 0.0864

b) 0.0384

c) 0.936

Step-by-step 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