Four people are chosen at random. What is the probability that:
No two of them have their birthday in the same month?
The first and second people chosen have their birthdays in the same month, given that there is some pair of people having their birthdays in the same month?
For the first question I think it's straight forward enough, P(no two of them have their birthday in the same month) = P(Everyone's birthday is the same) = $\frac {12}{12} *\frac {11}{12}*\frac {10}{12}*\frac {9}{12}$
For the second question I used Bayes' Theorem the set up I have is:
$P(1st \,2nd \,Same|P(Some\, pair\, same) = \frac {P(some \, pair\, same\,)|P(1st \, 2nd\, same)*P(1st\, 2nd\, same)}{P(Some\, pair\, same)} = \frac {1* (\frac{12}{12}*\frac{1}{12})}{\binom{4}{2} \frac{12}{12} \frac{1}{12} \frac{11}{12} \frac{11}{12}}$
The solution seems to suggest that P(Some pair having the same birthday) = $1- \frac {12}{12} *\frac {11}{12}*\frac {10}{12}*\frac {9}{12}$ which doesn't make any sense to me.
How can the complement of P(all birthdays are different) = P(some pair have same birthday). I thought the probability of some pair have the same birthday is in format like: AABC, ABCC, etc. Shouldn't the complement of P(all birthdays are different) means that Not all birthdays are the same meaning that you could have cases like AAAB, ABBB ?
It rather depends on whether having two pairs of shared months, or one triplet or one quadruplet count as having some pair having the same birthday month. In such cases there is indeed a pair with a shared month, but there are in fact several such pairs
If the question had been The first and second people chosen have their birthdays in the same month, given that there is exactly one pair of people having their birthdays in the same month? then the answer would be $\dfrac{1}{4 \choose 2}$
But if the question means The first and second people chosen have their birthdays in the same month, given that there is at least one pair of people having their birthdays in the same month? then the answer would be $\dfrac{\frac{12}{12}\times \frac{1}{12}}{ 1- \frac {12}{12} \times\frac {11}{12}\times\frac {10}{12}\times\frac {9}{12}}$