probability of having 2 boys given at least 1 is a boy I am in 12th grade and I can't really understand a question in conditional probability.
The question states

There is a family with 2 children either boys or girls where both are equally likely. Given that there is at least 1 boy find the probability of having 2 boys.

If we take the 2 cases, in the first case if there is 1 boy does it mean that there are 1 boy and 1 girl? Or does it mean that we have 1 boy and the 2nd may or may not be a boy?
 A: I can understand your confusion. Always write down the total sample space and sample space under condition. This is a question of conditional probability. The total sample space in this case is $\lbrace B, B \rbrace$, $\lbrace B, G \rbrace$, $\lbrace G, B \rbrace$,  $\lbrace G, G \rbrace$ each case with equal probability $1/4$.  When it says that there is "atleast one boy". Then your conditional sample space shrinks to  $\lbrace B, B \rbrace$, $\lbrace B, G \rbrace$, $\lbrace G, B \rbrace$,. Now  is asking the probability of having 2 boys, hence your favourable sample space is $\lbrace B, B \rbrace$. Hence the probability is = number of favourable sample space/ number of conditional sample space = 1/3.
A: The question means that there is at least one boy, second maybe a boy or a girl. You need to give the probability that both children are boys.
Without the information that at least one is a boy we would have four possible events
BB, BG, GB, GG. The probability of having two boys (BB) would then be 1/4 as each of the events has the same probability. But now we know that GG is not possible so we have only 3 possibilities so the answer is 1/3.
In greater detail:
Since the birth of boy or girl has the same probability Births in each order is given by
P(BB)=0.5*0.5
P(BG)=0.5*0.5
P(GB)=0.5*0.5
P(GG)=0.5*0.5
Applying Bayes theorem P(BB|one is boy)=$\frac{P(BB)}{P(BB)+P(GB)+P(BG)}=1/3$
