Respuesta :
Using the binomial distribution, it is found that there is a 0.3963 = 39.63% probability that he misses at least one.
For each free throw, there are only two possible outcomes. Either it is made, or it is missed. The probability of a free throw being made is independent of any other free throw, which means that the binomial distribution is used.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 90.4% probability of making a free throw, thus [tex]p = 0.904[/tex]
- 5 free throws, thus [tex]n = 5[/tex].
The probability of missing at least one is the probability of making less than 5, which is P(X < 5), which by:
[tex]P(X < 5) = 1 - P(X = 5)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{5,5}.(0.904)^{5}.(0.096)^{0} = 0.6037[/tex]
Then
[tex]P(X < 5) = 1 - P(X = 5) = 1 - 0.6037 = 0.3963[/tex]
0.3963 = 39.63% probability of missing at least one.
A similar problem is given at https://brainly.com/question/24756209
