A box contains 12 balls numbered 1 through 12. If 5 balls are selected one at a time from the box, without replacement, what is the probability that the largest number selected will be 9?
I want to just say $$\frac{9\times8\times7\times6\times5}{\binom{12}{5}}$$ but that is wrong and I don't know why.
Best Answer
You need to do the problem in two parts. First what is the probability that none of the ball chosen are numbered 10 through 12. This has probability $P=\frac{9\times 8\times 7\times 6 \times 5}{12 \times 11\times 10\times 9\times 8}$. Second the probability (under this condition) that a 9 has been chosen. To get this calculate the probability that a 9 was not chosen. This is $Q=\frac{8\times 7\times 6\times 5\times 4}{9\times 8\times 7\times 6\times 5}$. The final answer is $P(1-Q)$