Find the probability that six different person birthdays will fall in exactly two calendar months.

Hint: We can find the total number of the possible outcomes by taking the number of ways of having 6 person birthday in any of the 12 months. Then we can find the number of ways the birthday of six different people fall in exactly two months by taking the product of a number of ways of selecting 2 months from 12 months and the number of ways the 6 persons can have a birthday in exactly two months. Then we get the required probability by dividing the number of favorable outcomes by the total number of outcomes.

Complete step by step answer:

Each person can have a birthday in either of the 12 months of the year. So, each person has a possibility of 12. Then the possible ways of selecting any month out of 12 months is given by 12 raised to power 6, i.e. ${12^6}$ This becomes the denominator to find the probability.

Now we can find the numerator. We can take any two months out of the 12 months in a calendar. It is given by the combination${}^{12}{C_2}$. Now we have 6 people who need to have birthdays on either one of these two months. The number of ways of one person having a birthday in any of the 2 months is given by 2. Then the number of ways of 6 people having a birthday in any of the two months will be ${2^6}$ It is given that the birthday must fall in exactly 2 calendar months. So we must subtract the possibility of having 6 person birthday in the same month. It is given by 2. So the number of ways the birthday of six different people fall in exactly two months is given by,
${}^{12}{C_2} \times \left( {{2^6} - 2} \right)$.
Therefore the probability that six different person birthdays will fall in exactly two calendar months is given by the number of ways the birthday of six different people fall in exactly two months divided by the total number of ways of having a birthday in any of the 12 months.
$ \Rightarrow P = \dfrac{{{}^{12}{C_2} \times \left( {{2^6} - 2} \right)}}{{{{12}^6}}}$
We know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
$ \Rightarrow P = \dfrac{{12!}}{{2!\left( {10} \right)!}} \times \dfrac{{\left( {{2^6} - 2} \right)}}{{{{12}^6}}}$
On cancelling common terms we get,
$ \Rightarrow P = \dfrac{{12 \times 11}}{2} \times \dfrac{{2 \times \left( {{2^5} - 1} \right)}}{{{{12}^6}}}$
On further simplification, we get,
$ \Rightarrow P = \dfrac{{11 \times \left( {31} \right)}}{{{{12}^5}}}$
$ \Rightarrow$ P = $\dfrac{{341}}{{{{12}^5}}}$

Note: We use the concept of permutations and combinations to find a number of ways. In the step of finding the number of ways the birthday of six different people fall in exactly two months, we must consider the terms ‘exactly two months’ and subtract the possible number of ways of six persons having a birthday in the same month. We took the combinations for a number of ways of selecting 2 months from 12 because the order is not important. We must understand when to take the sum and product while finding a number of ways.