Respuesta :
2 and 1/2 because 2 shots would be 50% and 3 would be 75% so it would have to be 2 and 1/2 shots.
Using the binomial distribution, it is found that there is a 0.1296 = 12.96% probability that he makes all 4 shots.
-------------------------------
For each throw, the player either makes it or misses it. The probability of making each throw is the same, independent of other throws, thus, the binomial distribution is used to solve this question.
Binomial probability distribution:
The binomial probability is the probability of exactly x successes on n repeated trials, with X having two possible outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of a success on a single trial.
- Makes 60% of the shots, thus [tex]p = 0.6[/tex]
- 4 shots, thus [tex]n = 4[/tex].
- The probability of making all of them is P(X = 4), thus:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 4) = C_{4,4}.(0.6)^{4}.(0.4)^{0} = 0.1296[/tex]
0.1296 = 12.96% probability that he makes all 4 shots.
A similar problem is given at https://brainly.com/question/9000732
