I'm learning Probability and Statistics for the first time and there is a part where my professor said "it's done the same way, let's skip" but I don't understand how it's done the same way.
The first question goes like this:
A family has two children, and it is known that at least one is a
girl. What is the probability that both are girls, given this
information?
For this I understood that the all possible cases of having 2 children are $ \{GG, GB, BG, BB\} $. So when calculating the probability, it would be
$$ P(both \space girls|at \space least \space one \space girl) = \frac{P(both \space girls \cap at \space least \space one \space girl)}{ P(at \space least \space one \space girl)} $$
Since the cases of having at least one girl is $\{GG,GB,BG\}$ so $ P(at \space least \space one \space girl)$ should be $\frac 34 $.
And the cases of $ both \space girls \cap at \space least \space one \space girl $ is when all are girls, $\{GG\}$, so $ P(both \space girls \cap at \space least \space one \space girl) $ is $\frac{1}{4}$.
So I put all the values in, $$ P(both \space girls|at \space least \space one \space girl) = \frac{P(both \space girls \cap at \space least \space one \space girl)}{ P(at \space least \space one \space girl)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} $$
So the first question I understood, but when I tried to solve the second question the same way I got confused.
The second question goes like this:
A family has two children. You run into one of them and she is a girl.
What is the probability that both children are girls given this
information. Assuming that you are likely to run into either child,
and that which one you run into has nothing to do with gender.
I thought that $ P(random \space child\space is \space girl) $ will be $\frac{3}{4}$, as in the first question, because the cases that you randomly meet a girl is $\{GG,GB,GB\}$ out of $ \{GG, GB, BG, BB\} $.
And $P(both \space girls \cap random \space child\space is \space girl)$ would be $\{GG\}$ out of $ \{GG, GB, BG, BB\} $, $\frac{1}{4}$.
So the answer for this question would be (what I thought) was $$ P(both \space girls|random \space child\space is \space girl) = \frac{P(both \space girls \cap random \space child\space is \space girl)}{ P(random \space child\space is \space girl)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} $$
However, in the book it says that for events $G_1, G_2, G_3$ that elder, younger, and random child is a girl respectively, $P(G_1), P(G_2), P(G_3)$ are all $\frac{1}{2}$.
For $P(G_1), P(G_2)$ I understand because it would be cases of $\{GG,GB\}$ and $\{GG,BG\}$ out of $ \{GG, GB, BG, BB\} $.
But shouldn't it be $\frac{3}{4}$ for $P(G_3)$ because the possible cases are $ \{GG, GB, BG\} $ out of $ \{GG, GB, BG, BB\} $?
Sorry for the long question and bad formatting (it's my first time). But I really don't understand how this is supposed to work. What makes the two questions different that the answers are different? I must be thinking in some wrong way… Could you please help? Thank you in advance!!
Best Answer
$G_3$ is the event that you meet a random child of the family, and that one is a girl.
So the outcomes are described by family-type and which-child. Of the eight equiprobable outcomes, four result in meeting a girl. $$\rm \{\underline{GG1}, \underline{GG2}, \underline{GB1}, GB2, BG1, \underline{BG2}, BB1, BB2\} $$
So, of the four equiprobable outcomes where you meet a girl, two belong to a family with two girls. You might have met the younger or older sibling with equal probability, but whichever it was, the probability that the other child is a girl is $1/2$.
Notice how you do not meet a girl in two outcomes where the family does have exactly one girl, but you meet her sibling.
This highlights the difference in information given by the statements "at least one is a girl", and "this one is a girl"