Respuesta :
Answer:
30.91% probability that he makes exactly three of his next four free throws
Step-by-step explanation:
For each free throw, there are only two possible outcomes. Either he makes ir, or he does not. The probability of making a free throw is independent of other free throws. 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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Makes 56% of free throws.
So [tex]p = 0.56[/tex]
Assuming free throws are independent, the probability that he makes exactly three of his next four free throws is:
This is P(X = 3) when n = 4. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 3) = C_{4,3}.(0.56)^{3}.(0.44)^{1} = 0.3091[/tex]
30.91% probability that he makes exactly three of his next four free throws
