Respuesta :
Answer:
There is a 0.02% probability that at least two people have the same birthday.
Step-by-step explanation:
There are only two possible outcomes: Either the people do have the same birthday or they do not. So we use the binomial probability distribution.
Binomial probability
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}.\pi^{x}.(1-\pi)^{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 [tex]\pi[/tex] is the probability of X happening.
In this problem, we have that:
There are 365 days in a year, so the probability that a person has a birthdday on any given day is given by [tex]\pi = \frac{1}{365} = 0.0027[/tex].
Suppose there is a room with 9 people in it, find the probability that at least two people have the same birthday.
There are 9 people, so [tex]n = 9[/tex].
We also want to find [tex]P(X > 1)[/tex]
And
[tex]P(X \leq 1) + P(X > 1) = 1[/tex]
[tex]P(X > 1) = 1 - P(X \leq 1)[/tex]
We also have that:
[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]
[tex]P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}[/tex]
[tex]P(X = 0) = C_{9,0}.(0.0027)^{0}.(0.9973)^{9} = 0.9760[/tex]
[tex]P(X = 1) = C_{9,1}.(0.0027)^{1}.(0.9973)^{8} = 0.0238[/tex]
--------
[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.9760 + 0.0238 = 0.9998[/tex]
[tex]P(X > 1) = 1 - P(X \leq 1) = 1 - 0.9998 = 0.0002[/tex]
There is a 0.02% probability that at least two people have the same birthday.
