>>13168197Say there's no croaking, can you tell the probability of both being male?
Sample space = {MM, FM, FF}
The answer is not 1/3, you said they socialize at random, not in equal amounts of pairs of genders. The latter is still random, but is as likely as any other distribution. Saying it is random is as good as not giving any information at all.
The probability is just P(MM) or 1-P(FF)-P(FM).
Say now we tackle your croaking problem.
The event C (croaking) means there is at least one frog, C = MM union FM.
By sum theorem: P(C) = P(MM)+P(FM)-P(MM intersects FM) = P(MM)+P(FM).
You ask for P(MM|C), which by definition is
P(MM intersects C)/P(C)
= P(MM)/P(C)
= P(MM)/(P(MM)+P(FM))
That is the final answer, if you don't like it go read a fucking book on probability. For the sake of not being pedant, let's say OP wasn't a faggot and knew probability, then he would have told us the frogs socialize on equal amounts for each gender pair. In which case P(MM) = P(FM) 1/3 if we don't consider FM to be twice as likely as the others (being FM and MF different events, then P(MM) = 1/4).
So an answer would be 2/3, but using the other distribution 1/3.
