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September 29th, 2017, 06:46 PM   #1
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Help with probability!


I am currently struggling to answer the following example:

What is the minimum number of children (N) a couple should have in order to achieve both a male and female child with 99% probability? (Given the presumptions that the chances of having either a boy or girl is 50/50, and each sex at birth is independent of previous births)

Any help would be great!!
louise113 is offline  
September 29th, 2017, 07:00 PM   #2
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Let's just do this with coin flips or bitstrings. The probability of a string of $n$ 0's is $\frac{1}{2^n}$ and the probability of $n$ 1's is also $\frac{1}{2^n}$. So the probability that there's at least one of each is just $1 - \frac{1}{2^{n-1}}$. Now you just have to find the $n$ for which $1 - \frac{1}{2^{n-1}} \ge .99$.

Then $.01 \ge \frac{1}{2^{n-1}}$ and $n = 8$.
Maschke is offline  

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