Solution: 2015 Winter Final - 14

Author: Michiel Smid

Question

Let $S$ be a set of 100 integers; 30 of these are positive and the other 70 integers in $S$ are negative. We choose, uniformly at random, a 20-element subset of $S$. What is the probability that at least one of the elements in this subset is positive?

(a)

$1 - \left. {70 \choose 20} \middle/ {100 \choose 20} \right.$

(b)

$\left. {30 \choose 1} \middle/ {100 \choose 20} \right.$

(c)

$\left. {30 \choose 1}{70 \choose 19} \middle/ {100 \choose 20} \right.$

(d)

None of the above.

Solution

Let S be the set of all possible 20-element subsets of S.

We simply choose 20 elements from the 100 elements in $S$: $ \binom{100}{20} $

$ |S| = \binom{100}{20} $

Let B be the event that none of the elements in the subset is positive.

The number of ways to choose 20 negative elements from the 70 negative elements in $S$: $ \binom{70}{20} $

$ |B| = \binom{70}{20} $

$ Pr(B) = \frac{\binom{70}{20}}{\binom{100}{20}} $

Let A be the event that at least one of the elements in the subset is positive.

$ Pr(A) = 1 - Pr(B) $

$ Pr(A) = 1 - \frac{\binom{70}{20}}{\binom{100}{20}} $