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Math Help - Expected family size

  1. #1
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    Expected family size

    An event has probability p of success and q(=1-p) of failure. Independent trials are carried out until at least one success and one failure has occurred. Find the probability that r trials are necessary (r\geq2) and show that this probability equals (\frac{1}{2})^{r-1} when p=\frac{1}{2}.
    A couple decide that they will continue to have children until they have both a boy and girl or they have four children. Assuming boys and girls are equally likely to be born, what will be the expected size of the completed family.

    P(p)=p+pq+pq^2+pq^3+...+pq^{r-1}
    P(q)=q+qp+qp^2+qp^3+...+qp^{r-1}
    Sum of the general term = P(p)+P(q)=pq(p^{r-2}+q^{r-2})
    when p=\frac{1}{2} \rightarrow q=\frac{1}{2}
    \frac{1}{4}[(\frac{1}{2})^{r-2}+(\frac{1}{2})^{r-2}]
    =(\frac{1}{2})^r+(\frac{1}{2})^r
    =2(\frac{1}{2})^r
    =(\frac{1}{2})^{r-1}
    So I have gotten these parts right, but about the last part, how do I do it?
    Because I suppose I use (\frac{1}{2})^{r-1} to find the probability of having a boy and girl after r trials. So I find the value when r is 2, 3, and 4? and then each of those multiply by the number of children?
    Thanks
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  2. #2
    MHF Contributor matheagle's Avatar
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    There are only three possibilities.
    Define the random variable X=number of children.

    P(X=2)=P(\{{\rm boy, girl\} \quad or \quad\{ girl,boy}\})=2pq

    Next obtain P(X=3) and P(X=4).
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  3. #3
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    Quote Originally Posted by matheagle View Post
    There are only three possibilities.
    Define the random variable X=number of children.

    P(X=2)=P(\{{\rm boy, girl\} \quad or \quad\{ girl,boy}\})=2pq

    Next obtain P(X=3) and P(X=4).
    X? so X is the number of children or in other words r in (\frac{1}{2})^{r-1}?
    thanks

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