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Suppose that a family has exactly n children (n ≥ 2). Assume that the probability that any child will be a girl is 1/2 and that all births are independent. Given that the family has at least one girl, determine the probability that the family has at least one boy.

This is how I went about it:

Since we know that the family has at least one girl, we now have to determine the probability that the family has at least one boy from a pool of n-1 children.

Thus, probability is equal to:

1-Probability of having no boys

=1-(1/2)^(n-1)

But the answer is (1-(1/2)^(n-1)) / (1-(1/2)^(n))

The structure of the answer suggests that Bayes theorem is being used in the correct answer, but even the Bayes theorem should lead to my answer. I do not know where I am going wrong?

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    $\begingroup$ Notice the difference between the family having at least one girl, and the family's oldest child being a girl. $\endgroup$
    – N.U.
    May 18 '13 at 13:02
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Define $Y$ to be the numbe of boys, $X$ the number of girls. You need: $$ P(Y \geq 1 |X \geq 1)=1-P(Y=0|X \geq 1) = 1- \frac{P(Y=0|X \geq 1)P(X \geq 1)}{P(X \geq 1)}\\ =1 -\frac{P(Y=0 \cap X \geq 1)}{1-P(X=0)}=1-\frac{\frac{1}{2^n}}{1-\frac{1}{2^n}}=\frac{1-\frac{1}{2^{n-1}}}{1-\frac{1}{2^n}} $$

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  • $\begingroup$ thanks a lot Alex and N.U.! $\endgroup$
    – user75001
    May 18 '13 at 13:07
  • $\begingroup$ You are welcome. Feel free to upvote answers you appreciate. $\endgroup$
    – Alex
    May 18 '13 at 14:06
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    $\begingroup$ vote up requires more than 15 reputation. Will do so when I gain the required reputation $\endgroup$
    – user75001
    May 18 '13 at 14:35
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    $\begingroup$ Nah, minor error. I make far worse! I only corrected it because some readers might be confused! $\endgroup$ May 18 '13 at 15:36

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