I am looking for the probability of Pc(B) where the event of B={no two people are born in the same month} and event C= {exactly three people were born in the summer of june, july august} and there are 9 people involved.
for
P(B) i got =1/12 * 1/11 * 1/10 * 1/9 * 1/8 * 1/7 * 1/6 * 1/5 * 1/4 =1/79833600
P(C)=(9 choose 3)((3/12)^3)((9/12)^6)=(15309/65536)
so
Pc(B)=P(C|B)=P(B intersection C)/P(C) = (P(B)*P(C))/P(C)
=((1/79833600) * (15309/65536))/(15309/65536)
Is my logic correct? if not where did i go wrong
Detailed: You have nine borns which exactly three of them are in the summer. This means you can not have all states of event $B$ with $C$(some of them will be ineligible). So they aren't independent. Now we have:
$$P(B \cap C) = (\frac{3\times2\times1}{12^3})(\frac{9\times8\times\ldots4}{12^6})$$ $$P(C) = (\frac{3}{12})^3(\frac{9}{12})^6$$ $$P(C|B)=\frac{P(B \cap C)}{P(C)}$$ Previous pervious answer was wrong:
Your events aren't independent. So $P(B \cap C) \ne P(B)*P(C)$. For instead you have $P(B \cap C) = (\frac{1}{3} \frac{1}{2} \frac{1}{1})(\frac{1}{9}\frac{1}{8}\frac{1}{7}\frac{1}{6}\frac{1}{5}\frac{1}{4})$