Respuesta :
Answer:
a) There is a 63.02% probability that no one has done an one time fling.
b) There is a 36.98% probability that at least one person has done a one time fling
c) There is a 99.15% pprobability that no more than two people have done a one time fling.
Step-by-step explanation:
For each adult, there are only two possible outcomes. Either they have done an one time fling, or they have not. This means that we use the binomial probability distribution to solve this problem.
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 combinatios 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.
In this problem we have that:
[tex]n = 9, p = 0.05[/tex]
A. no one has done a one time fling
This is P(X = 0).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{9,0}.(0.05)^{0}.(0.95)^{9} = 0.6302[/tex]
There is a 63.02% probability that no one has done an one time fling.
B. at least one person has done a one time fling
Either no one has done a one time fling, or at least one person has. The sum of the probabilities of these events is decimal 1.
So
[tex]P(X = 0) + P(X \geq 1) = 1[/tex]
[tex]P(X \geq 1) = 1 - `(X = 0) = 1 - 0.6302 = 0.3698[/tex]
There is a 36.98% probability that at least one person has done a one time fling
c. no more than two people have done a one time fling
This is
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
[tex]P(X = 1) = C_{9,1}.(0.05)^{1}.(0.95)^{8} = 0.2985[/tex]
[tex]P(X = 2) = C_{9,2}.(0.05)^{2}.(0.95)^{7} = 0.0628[/tex]
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.6302 + 0.2985 + 0.0628 = 0.9915[/tex]
There is a 99.15% pprobability that no more than two people have done a one time fling.
