# Math Help - More expected value

1. ## More expected value

Hi!

A family has children until it has a boy or until it has three children, whichever comes first. Assume the child is a boy with probability $\frac{1}{2}$. Find the expected number of boys in this family, and the expected number of girls.

Let $X$ denote the number of boys in the family.
Let $Y$ denote the number of girls in the family.

We are seeking $E(X)$ and $E(Y)$.

There are the following possible "child outcomes": G = girl B = boy

$\Omega = \left(GGG,GGB,GB,B\right)$

It is not possible to have for example BBG, since the family stops having children when they have gotten a boy.
Let $F$ be any of the events in $\Omega$.
Hence,

$E(X)=\displaystyle \sum_{\Omega} E(X|F)P(F) = \frac{3}{4}$

$E(Y)=\displaystyle \sum_{\Omega} E(Y|F)P(F) = \frac{3}{2}$

Did I go wrong somewhere, if so, what is the error?

Thanks!

2. Originally Posted by Twig
Hi!

A family has children until it has a boy or until it has three children, whichever comes first. Assume the child is a boy with probability $\frac{1}{2}$. Find the expected number of boys in this family, and the expected number of girls.

Let $X$ denote the number of boys in the family.
Let $Y$ denote the number of girls in the family.

We are seeking $E(X)$ and $E(Y)$.

There are the following possible "child outcomes": G = girl B = boy

$\Omega = \left(GGG,GGB,GB,B\right)$

It is not possible to have for example BBG, since the family stops having children when they have gotten a boy.
Let $F$ be any of the events in $\Omega$.
Hence,

$E(X)=\displaystyle \sum_{\Omega} E(X|F)P(F) = \frac{3}{4}$

$E(Y)=\displaystyle \sum_{\Omega} E(Y|F)P(F) = \frac{3}{2}$

Did I go wrong somewhere, if so, what is the error?

Thanks!
i think, you better construct the probability distribution first before find the expected value....
probability distibution from
$\Omega = \left(GGG,GGB,GB,B\right)$
and Let $X$ denote the number of boys in the family.

the probability distribution
X P(X=x)
0 1/8
1 7/8

you can find expected value easily from the probability distribution table...
$E(X)=\sum(Xi Pi)$

......hope it help..........

3. Hi

Thanks you´re right.