1
$\begingroup$

Among 5 people, what is the probability that exactly 2 of them are born in the same month.

For "at least $2$ in the same month", my answer would be $$1 - \frac{12\cdot 11\cdot 10\cdot 9\cdot 8}{12^5}$$ the complement of probability that all months are distinct. How to deal with "exactly 2"?

(Assuming the birthdays are independent, and equally distributed among the months.)

$\endgroup$
  • $\begingroup$ Could you please show your efforts? $\endgroup$ – Pavel Yudaev Nov 14 '15 at 19:27
  • 1
    $\begingroup$ The question can not be answered without making assumptions about the probabilities. The question probably wants you to assume equal probability for birth in each month. That assumption though is not true. This is a good example of a very poorly phrased question. $\endgroup$ – Ittay Weiss Nov 14 '15 at 19:30
  • 1
    $\begingroup$ Its okay to down vote problems, but it would be nice to leave a note or write a reason. $\endgroup$ – alex alexeq Nov 14 '15 at 19:44
1
$\begingroup$

It is tacitly assumed that there are $12^5$ equiprobable cases. We now have to count the favorable cases. There are ${5\choose2}=10$ ways to select the two people having birthday in the same month, and $12$ ways to select that month. There are three people left, to which we can assign a month each in $11\cdot10\cdot 9$ ways. The probability in question therefore comes to $$p={10\cdot 12\cdot 11\cdot 10\cdot 9\over 12^5}={275\over576}\doteq0.47743\ .$$

$\endgroup$
1
$\begingroup$

We'll consider 12 different cases:

(1) Exactly two people are born in Jan and the remaining three are born in different months.

(2) Exactly two people are born in Feb and the remaining three are born in different months.

... and so on. Notice that the cases are mutually excluding, i.e., an event were exactly two people are born the same month can't be in two cases at the same time. Hence, the sought probability would be the sum of probabilities of each case.

What's the probability that two people are born in Jan and the rest during different months? There are ${5\choose2}$ ways to pick 2 people out of 5. There is a $\frac1{12}$ chance that a person is born in Jan (since we're assuming births are uniformly distributed), so there is a $(\frac1{12})^2$ that two people are born the same month. In this case, is Jan. Now we want the third person to be born in any month other than Jan. There are 11 months to choose from. Once chosen, the fouth person must be born in a different month than Jan and the one where the third person was born. Thus, there are 10 months to pick from. And so on.

Hence, it would be a $p={5\choose2}(\frac1{12})^2\cdot \frac{11}{12}\cdot\frac{10}{12}\cdot\frac9{12}$ chance that exactly two people are born in Jan and the remaining three are born in different months.

Since the months and births are independent from each other, the 11 remaining cases are exactly the same.

Then your answer would be $12p$

$\endgroup$
  • $\begingroup$ minus the probability that two of the remaining are both born in july... $\endgroup$ – djechlin Nov 14 '15 at 20:01
  • $\begingroup$ My reasoning is that I pick two out of five people, ${5\choose2}$, then there is a prob. of $\frac12^2$ that both are born the same month. The third is born in a different month and there are 11 months remaining to choose from. The 4th is born in different month, and there a 10 months to choose from and so on... Am I right? $\endgroup$ – EA304GT Nov 14 '15 at 20:06
  • $\begingroup$ Sorry, you're totally right. This answer still makes me nervous because there's some reasoning that's not spelled out, and usually is fallacious, and happens to be correct here, in particular, because your test event is rigged to be independent over the months. Which is a good strategy, but the reader could definitely get the impression that the independence is not important. $\endgroup$ – djechlin Nov 14 '15 at 20:18
  • $\begingroup$ fwiw I verified the "is january" part experimentally as a sanity check. $\endgroup$ – djechlin Nov 14 '15 at 20:19
  • $\begingroup$ Can you remove the (-1) vote then, please? I'll edit my answer to make my reasoning clearer. $\endgroup$ – EA304GT Nov 14 '15 at 20:21
0
$\begingroup$

Hint : Fix two persons and a month. ($120$ possibilities so far). Now, choose three other months ($165$ possibilities). In each case, you have $6$ possibilities to distribute the three months over the three remaining persons.

Do not bother Ittay's comment. Even, if the probabilities are not equal, it is OK to ask for the probability assuming equality.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.