A basketball player has made 70% 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) Misses for the first time on his fifth attempt b) Makes his first basket on his fourth shot c) Makes his first basket on one of his first 3 shots a) The probability that in tonight's game the basketball player misses for the first time on his fifth attempt is

a) 0.07203

b) 0.0189

c) 0.973

Step-by-step explanation:

Probability of making one shot, for the player = 70% = 0.70

Probability of NOT making a shot = 1 - 0.70 = 0.30

Assuming that the shots are independent

a) Misses for the first time on his fifth attempt

For him to miss on his for the first time on his fifth attempt, he has to make the first four shots

The required probability = (0.7)⁴(0.3) = 0.07203

b) Makes his first basket on his fourth shot

For him to make his first basket on the fourth shot, he has to miss the first three shots.

Required probability = (0.3)³(0.7) = 0.0189

c) Makes his first basket on one of his first 3 shots

This is a sum of probabilities; that is, he could make the first basket on the first shot, or miss on the first shot and make the second shot or miss on the first two shots and make the basket on the third shot.

Required probability

= (0.70) + (0.3×0.7) + (0.3×0.3×0.7)

= 0.70 + 0.21 + 0.063 = 0.973

Hope this Helps!!!

what do you need help with?

Step-by-step explanation: