An urn contains balls numbered $1, \ldots, 10$. Five balls are drawn without replacement. What is the probability that the second largest of the five numbers drawn will be 8?
I believe the number of possibilities are 10 * 9 * 8 * 7 * 6 but that is where I get stuck.
Any help would be appreciated.
Best Answer
Break it down into two parts, the probability that 8 and 10 are chosen, and 9 isn't; and the probability that 8 and 9 are chosen and 10 isn't. If 8, 9, and 10 are removed and we choose 3 from the remaining 7, we get
$$ {7 \choose 3} $$
for the number of ways of 8 being the second largest with 10 as the largest, and also with 9 as the largest. So the number of ways of choosing 5 balls with 8 as the second largest is
$$ 2*{7 \choose 3} $$
The number of ways of choosing 5 balls from 10 is
$$ {10 \choose 5} $$
So the answer is
$$ \frac{2*{7 \choose 3}}{{10 \choose 5}} $$