Three distinct integers are chosen at random from the first 20 positive integers. Compute the probability that: (a) their sum is even; (b) their product is even.
+128473 Here's my best attempt
We have C (20, 3) = 1140 possible different sets of 3 integers
(a) the sum will be even when we have
even + even + even or
even + odd + odd
In the first case we want to choose any 3 of the 10 even numbers = C (10, 3) = 120
In the second case we want to choose any 1 of 10 evens = C (10, 1) and any 2 of the 10 odds = C(10,2) = 45
So the probability is
[ C (10,3) + C(10,1) * C(10,2) ] 120 + 10 * 45 570
____________________________ = _____________ = ______ = .5
C (20, 3) 1140 1140
(b) the only way we can have an odd product is when all three integers are odd.....there are C (10,3) = 120 sets of these
So.....the probability that we have an even product =
1 - 120 / 1140 =
[ 1140 -120 ] / 1140 =
1020 / 1140 =
17 /19
