Conditional Probability with boys and girls

Question

Assume that a newborn baby is a girl with probability $p$ and a boy with probability $1-p$. Also assume that the genders of dierent newborns are independent of each other. Consider a person who has two children. Define the following two events:

A = "both children are girls"

B = "at least one of the children is a girl"

What is the conditional probability $Pr(A \vert B)$?

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My attempt is to use the conditional probability formula $P(A\vert B) = \frac{P(A\cap B)}{P(B)}$

$S = \{ BB, GG, GB, BG \}$

$P(A) = p\cdot p$ since for both children to be girls

$P(B) = 2\cdot\{(1-p)\cdot p\} + (p)^2 = 2p - p^2$ since the probability for at least one girl means the probability $GG$ plus $GB$ and $BG$, with NO $BB$

$P(A\cap B) = ?$ How do I get the intersection? I know the only intersection is $GG$ which is $\frac{1}{4}$ but I need to express it as a probability.

My guess is that the intersection is $p^2$ since that is the $P(A)$, but I'm not sure, can some one convince me if I'm right?

$P(A\vert B) = \frac{p^2}{2p-p2} =\frac{p}{2-p}$

Is this correct?

Answer 1

You are correct. The intersection is $GG$ as you have noticed. By the product law, the probability of $GG$ is:

$$P(GG)=p \times p = p^2$$

So based on the conditional probability formula:

$$P(A|B) = {P(A \cap B) \over P(B)} = {P(GG) \over 2p - p^2} = {p^2 \over 2p - p^2} = {p \over 2 - p}$$

In this case, since $B$ is a superset of $A$, the intersection $B \cap A$ is simply $A$. In other questions, you may need to calculate the probability of the intersection separately, as it may not always be equivalent to one of the sets you are given. Treat the intersection as another set and find its probability the same way you would for the sets $A$, $B$, etc. that you are given.

Answer 2

$P(B) = 1 - P(B') = 1 - (1 - p)^2 = 2p -p^2$, and $P(A \cap B) = P(A) = p^2$ since $A \subseteq B$ so you get the answer.