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A problem states:

There are 8 boys and 12 girls in a chess club. Two boys and two girls are to be selected at random to represent the club at a tournament. What is the probability that Jacob and Emily will both be chosen?

I calculate this as follows:

Probability that Jacob is chosen is $\frac{7}{28}$
Probability that Emily is chosen is $\frac{11}{66}$
Probability that both Jacob and Emily are chosen is $\frac{7}{28} \cdot \frac{11}{66} = \frac{1}{24}$

To answer the question: "What is the probability of neither Emily nor Jacob being chosen?" one would calculate the "$1-P$" of each of the individual probabilities and multiply those, so:
Probability that Jacob is not chosen = $1-\frac{7}{28}=\frac{3}{4}$
Probability that Emily is not chosen = $1-\frac{11}{66} =\frac{5}{6}$
Probability that neither Emily nor Jacob are chosen = $\frac{3}{4}\cdot\frac{5}{6}=\frac{15}{24}$

But it seems that another way to answer it would be to say that the question is equivalent to saying "$1-{}$the probability of both Emily and Jacob being chosen", or $1-\frac{1}{24}$, but that gives a different answer ($\frac{23}{24})$ than the above.

What am I missing?

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up vote 7 down vote accepted

The opposite of "neither Emily nor Jacob being chosen" is not "both Emily and Jacob being chosen". You must also account for the possibility that only one of them is chosen.

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@Nick: you should be able to calculate the probability that (Jacob is chosen and Emily is not) and the probability that (Emily is chosen and Jacob is not). Adding the four probabilities should give $1$. – Ross Millikan Aug 20 '11 at 15:37

You can also just calculate this using binomial coefficients. Out of all choices of two students, there is one way to choose both Jacob and Emily out of the group of 20. The total number of ways to choose 2 students is $\binom{20}{2}$ and so the probability that both Emily & Jacob are chosen is: $\frac{1}{\binom{20}{2}}$

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