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.
$\endgroup$ 23 Answers
$\begingroup$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
$$ \frac{2*{7 \choose 3}}{{10 \choose 5}} $$
$\endgroup$ $\begingroup$Here's a hint: This would be equivalent to asking what the probability of (drawing a 10 OR a 9, but not both) AND (drawing an 8).
Try using those conditions to solve the problem
$\endgroup$ $\begingroup$Another way of solving it is to use multivariate hypergeometric distribution:
Let us define A= 8, B = 9 or 10, C is three numbers from {1,2,..7}
Then $P(A = 1, B = 1, C = 3) = \dfrac{ {m1\choose a}{m2\choose b}{m3\choose c}}{{M\choose y}}$
In our problem, $m1 =1, m2 = 2, m3 = 7, a = 1, b = 1, c = 3, M = m1+m2+m3 = 10 , y = a+b+c = 5$
$$= \dfrac{ {1\choose 1}{2\choose 1}{7\choose 3}}{{10\choose 5}}$$
That will be solution.
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