# Probability that 3 people are born on different months from each other

torquerotates

## Homework Statement

find the probability that 3 people are born on different months from each other

## The Attempt at a Solution

So here's my reasoning. There are 3 people X1, X2, X3. Let (Xi,Xj) denote the pair have different birthmonths. P[(X1,X2) and (X2,X3) and (X3,X1)]=P(X1,X2)*P(X2,X3)*P(X3,X1)

Well P(Xi,Xj)=11/12
thus P[(X1,X2) and (X2,X3) and (X3,X1)]=P(X1,X2)*P(X2,X3)*P(X3,X1)=(11/12)^3

But that is not the answer. My book said the answer is (11/12)*(10/12).

I dont see what is wrong with my reasoning. for the 3 to have different birthmonths, any two would have different birthmonths. X1 different from X2, X2 from X3, X3 from X1.

Do you mean that no two are born in the same month, or that it is not the case that the three are born in the same month, and that you are assuming that the probability of being born in any month is equally-likely ?

torquerotates
Sorry. The question mentions that no two people are born on the same month. Also, the probability being born in any month is equality likely.

Sorry, my quote function does not work. Notice that X1,X2 may have been born in different months, and same for X2,X3, but it is still possible for X1,X3 to have been born in the same month as each other.

Basically: how many total choices of months are there for the 3 people to have been born in? And, out of all those choices, how many satisfy your condition ? One way of seeing the last is that, once X1, say has "chosen" a month to be born in, then you don't consider that choice for X2, and so on. Try a tree in which the root is the month X1 is born in, and where that month is no longer considered for X2, etc.

torquerotates
Notice that X1,X2 may have been born in different months, and same for X2,X3, but it is still possible for X1,X3 to have been born in the same month as each other.
.

I think i accounted for X1 and X3 when i set up the equation P[(X1,X2) and (X2,X3) and (X3,X1)]=P(X1,X2)*P(X2,X3)*P(X3,X1). There are only 3 combinations where any two cannot be equal to each other.

torquerotates
Basically: how many total choices of months are there for the 3 people to have been born in? And, out of all those choices, how many satisfy your condition ? One way of seeing the last is that, once X1, say has "chosen" a month to be born in, then you don't consider that choice for X2, and so on. Try a tree in which the root is the month X1 is born in, and where that month is no longer considered for X2, etc.

Thanks. That makes sense. I didn't think of approaching the problem this way at first. The most intuitive way for me to approach counting isn't to take up one slot at a time but to consider the whole picture like looking at the set of 3 and then taking 3C2.