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?