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) 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

Answer 1

Answer:

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.

So

$p = 0.4P(X = 5)$

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

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

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.

So

So

$p = 0.6P(X = 0)$

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

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

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 2

Answer:

Step-by-step explanation:

The probability that the basket player made a foul shot is 60% which is 0.60

Then the probability of good shot is 1 - 0.60 = 0.40

P = 0.40

a) the probability that the basket player misses for the first time on his sixth attempt ​is

P (first time on his sixth attempt) = (1 - P)⁵ (P)

= (1 - 0.4)⁵(0.4)

= (0.6)⁵(0.4)

= 0.07776 * 0.4

= 0.031104

≅ 0.0311

The probability that​  the basketball player misses for the first time on his sixth attempt is 0.0311

b) P(first basket on his fifth shot) = (1 - P)³ (P)

= (1 - 0.4)⁴(0.4)

= (0.60)⁴(0.4)

= 0.0518

c) The probability of making his first basket in first shot is 0.6

and the  probability of making his first basket in second shot is

0.6 * 0.4 = 0.24

the  probability of making his first basket in third shot is

0.6 * 0.4² = 0.096

So, the probability that the player makes his first basket on one of his first 3 shots is

= 0.6 + 0.24 + 0.096

= 0.936

Thus, the probability that in​ tonight's game the basketball player makes his first basket on one of his first 3 shots is 0.936