Mathematics, 19.05.2020 18:09 sammy16940

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) Misses for the first time on his sixth attempt b) Makes his first basket on his fifth 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 sixth attempt is

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Answer from: kdlishs

a) The probability that in tonight's game the basketball player misses for the first time on his sixth attempt is 0.0311 = 3.11%.

b) The probability that in tonight's game the basketball player makes his first basket on his fifth shot is 0.0154 = 1.54%.

c) The probability that in tonight's game the basketball player makes his first basket on one of his first 3 shots is 0.936 = 93.6%.

Step-by-step explanation:

For each shot, there are only two possible outcomes. Either the player makes it, or he does not. The probability of making a shot is independent of other shots. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

A basketball player has made 60% of his foul shots during the season.

This means that p = 0.6

a) Misses for the first time on his sixth attempt

Makes the first five, which is P(X = 5) when n = 5.

Misses the sixth, with probability = 1-0.6 = 0.4.

p = 0.4P(X = 5)

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$p = 0.4P(X = 5) = 0.4*(C_{5,5}.(0.6)^{5}.(0.4)^{0}) = 0.0311$

The probability that in tonight's game the basketball player misses for the first time on his sixth attempt is 0.0311 = 3.11%.

b) Makes his first basket on his fifth shot

Misses the first four, which is P(X = 0) when n = 4.

Makes the fifth, with a probability of 0.6.

p = 0.6P(X = 0)

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$p = 0.6P(X = 0) = 0.6*(C_{4,0}.(0.6)^{0}.(0.4)^{4}) = 0.0154$

The probability that in tonight's game the basketball player makes his first basket on his fifth shot is 0.0154 = 1.54%.

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

Either he makes his first basket on one of his first 3 shots, or he misses all of them. The sum of these probabilities is decimal 1.

Misses the first three:

P(X = 0) when n = 3. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.6)^{0}.(0.4)^{3} = 0.064$

Makes on one of his first three:

1 - 0.064 = 0.936

The probability that in tonight's game the basketball player makes his first basket on one of his first 3 shots is 0.936 = 93.6%.

Answer from: Quest

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Answer from: Quest

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