Answer:
0.0031 = 0.31% probability that exactly 7 out of 10 individuals will visit the Statue of Liberty
Step-by-step explanation:
For each individual, there are only two possible outcomes. Either they visit the Statue of the Liberty, or they do not. The probability of an individual visiting the statue is independent of any other individual. This means that the binomial probability distribution is used 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.
The sample probability of an individual traveling to the statue is 25%
This means that [tex]p = 0.25[/tex]
The probability that exactly 7 out of 10 individuals will visit the Statue of Liberty is
This is [tex]P(X = 7)[/tex] when [tex]n = 10[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 7) = C_{10,7}.(0.25)^{7}.(0.75)^{3} = 0.0031[/tex]
0.0031 = 0.31% probability that exactly 7 out of 10 individuals will visit the Statue of Liberty
