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.
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."
adds no extra information to
"The older child is a girl."
because it adds another possibility
" The younger child is also a girl."
by which the circle of possibilities gets larger.
Information is added if the circle of possibilities shrinks.