Find the probability of two of them having the same birthday

Question

If six people are selected at random. Find the probability of two of them having the same birthday (day and month) is 365/365*364/365

Answer 1

There's 4 elements we need to solve:

 

[Number of days in a year] *

[Combinations of 2 people from a group of 6] *

[Probability of having a birthday on a day]^[number of people having a birthday that day] *

[Probability of not having a birthday on that day]^[number of people not having a birthday that day]

 

365C1 * 6C2 * (1/365)^2 * (364/365)^4

= .0407

 

This is a binomial distribution of the number of ways two people can have a birthday on the same day, but since there are 365 days in a year, you need to multiply that distribution by 365.

Answer 2

A really neat thing about probabilities of any event is that they must add to 1 (or 100%).  The probability of any single event is (#success/#total).

 

So, the way to determine the probability of 2 (or more) people sharing the same birthday is to calculate the probability that NOBODY shares a birthday, and then subtract it from 1.

 

When you determine this probability of multiple events, simply multiply the probability of each event.

 

The first person has a birthday.  So, he has 365 chances of a unique birthday, out of 365 total days possible.

P(1st)=365/365=1

 

The 2nd person has a birthday, but it must be unique from the first person.  

P(2nd)=364/365=0.99726

 

The 3rd person must be unique from the other two.

P(3rd)=363/365=0.99452

 

P(4th)=362/365=0.99178

 

P(5th)=361/365=0.98904

 

P(6th)=360/365=0.98630

 

Multiply all of the probabilities, and you get 0.9595 as the probability that ALL are unique.

 

Subtract from one to determine the probability that they are NOT all unique and you get 1-0.9595=0.0405 or about 4%.