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A basketball player has made 80​% 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 fourth fourth attempt ​b) Makes his first basket on his fourth fourth shot ​c) Makes his first basket on one of his first 3 shots

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Final answer:

a) The probability of missing for the first time on the fourth attempt is 0.16%. b) The probability of making the first basket on the fourth shot is 0.8%. c) The probability of making the first basket on one of the first 3 shots is 99.2%.

Step-by-step explanation:

a) To find the probability that the basketball player misses for the first time on his fourth attempt, we need to consider that each shot is independent. Since he has made 80% of his shots, the probability of making a shot is 0.8. Therefore, the probability of missing a shot is 1 - 0.8 = 0.2. Since the shots are independent, the probability of missing on the fourth attempt is 0.2 raised to the power of 4 = 0.0016, or 0.16%.

b) To find the probability that the basketball player makes his first basket on his fourth shot, we use a similar approach. The probability of making a shot is 0.8, and the probability of missing a shot is 0.2. The probability of making a shot on the fourth attempt is then 0.2 raised to the power of 3, multiplied by 0.8 = 0.2 * 0.2 * 0.2 * 0.8 = 0.008, or 0.8%.

c) To find the probability that the basketball player makes his first basket on one of his first 3 shots, we need to find the complementary probability. The probability of not making a basket on the first 3 shots is 0.2 raised to the power of 3 = 0.008. Therefore, the probability of making a basket on one of the first 3 shots is 1 - 0.008 = 0.992, or 99.2%.

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