# Three people are picked at random from a group of 365. What is the probability they are all born in the same month.

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The sum of the probability of an event A taking place and the probability of the event not taking place is equal to 1.

When three people are picked from a group, the probability that they are **not born in the same month is equal to (11/12)*(10/12)**. This is the case as irrespective of which month the first person is born in, the birthday of the second can be in the remaining 11 months and the third can be born in any of the remaining 10 months.

This gives the probability that the three people are born in the same month as 1 - (11/12)*(10/12) = 17/72

There is a probability of 17/72 that the three are born in the same month.

Sorry for a **small error** made above. When the three people are not born in the same month two of them can be born in the same month different from that in which the first is born. This gives the probability that the three are not born in the same month as (11/12)*(11/12) = 121/144

**The probability that the three are born in the same month is 1 - 121/144 = 23/144**

What is the probability that three people chosen randomly will have the same birth month?

(1) Assuming that births are randomly distributed between the months:

Choose one of the three people. Then the next person has a 1/12 chance of sharing a birth month, and the third person will also have a 1/12 chance of sharing the birth month. Thus the probability is 1/144 or approximately .007

(2) Another way to look at this with the above assumption. There are 12x12x12=1728 ways to group three objects with 12 possible attributes. There are 12 ways that the three will have the same attribute (e.g. all jan, all feb, etc)

So the probability is 12/1728 or approximately .007

(3) If we assume that birthdays are evenly distributed throughout the year: (Note that this eliminates the problem of different months having different numbers of days)

The probability of being born in Jan is 31/365. So the probability that all three are born in Jan is `(31/365)(31/365)(31/365)=(31/365)^3 `

Similarly,for Feb (ignoring leap years) we haveĀ `(28/365)^3 `

We can do the same for all 12 months -- since each event is mutually exclusive we can add the probabilities:

`(31/365)^3+(28/365)^3+(31/365)^3+(30/365)^3+(31/365)^3+(30/365)^3 `

`+(31/365)^3+(31/365)^3+(30/365)^3+(31/365)^3+(30/365)^3+(31/365)^3 `

`~~.007 `

These all ignore the fact that birthdays are not perfectly randomly distributed through the year.

1-(1/12*11/12*11/12)