Conditional probability involving boy girl scenario

Consider a family that has two children. We are interested in the children's genders. Our sample space is S={(G,G),(G,B),(B,G),(B,B)} (G for Girl, B for Boy). Also assume that all four possible outcomes are equally likely.

A) What is the probability that both children are girls given that the first child is a girl? B) We ask the father: "Do you have at least one daughter?" He responds "Yes!" Given this extra information, what is the probability that both children are girls?

for the first question i initially thought it was a 1/2 but then i realised a girl was already present so excluded BB. And i cam to 2/3 idk if thats right or if its just 1/3.

And i do not understand the conditional given that part as i have never been taught to find the cardinality of a sample or something and exam questions were straightforward in the sense of not needing to find (AnB) by seeing what is common in both sets etc. Now in University the lecturer is saying different. i have again come to the conclusion of 1/3 which is clearly wrong. Many thanks guys.

• For the first one, we already know the first child is a girl, so the only two possible cases are $GB$ and $GG$. Hence, your first thought of $\frac 12$ is correct. – lulu Nov 29 '17 at 17:40
• For the second one, why do you say $\frac13$ is "clearly incorrect"? – lulu Nov 29 '17 at 17:41
• Possible duplicate of Variant of boy or girl paradox – Karn Watcharasupat Nov 29 '17 at 17:49
• This question has been asked before here. – Karn Watcharasupat Nov 29 '17 at 17:49

We want to calculate certain conditional probabilities. We'll do that blindly based on the definitions.

1 $$P(\{(G,G)\} \ \mid \ \{(G,G),(G,B)\})=\frac{P(\{(G,G)\}\cap\{(G,G),(G,B)\})}{P(\{(G,G),(G,B)\})}=$$ $$=\frac{P(\{(G,G)\})}{P(\{(G,G),(G,B)\})}=\frac{\frac14}{\frac24}=\frac12.$$

In the second case

2

$$P (\{(G,G)\}\ \mid \{(G,G),(G,B),(B,G)\})=\frac {P (\{(G,G)\})}{P( \{(G,G),(G,B),(B,G)\})}=\frac13.$$

Note.

"At least one of the children is a girl."