It seems to me that there is a mistake in the official solution to the following problem offered on Alcumus:
Four prime numbers are randomly selected without replacement from the
first ten prime numbers. What is the probability that the sum of the
four selected numbers is odd? Express your answer as a common
fraction.
The official solution reads as follows:
The first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.
The sum of four of these numbers is only odd if 2, the only even
number on the list, is among them because the sum of four odd numbers
is even. Once 2 is picked, there are $\binom{9}{3}=\frac{9!}{3!6!}=84$
ways to pick three numbers from the remaining nine. The total number
of ways to select four prime numbers from the ten is
$\binom{10}{4}=\frac{10!}{4!6!}=210$. Therefore, the probability that
the sum of the four selected numbers is odd is
$\frac{84}{210}=\mathbf{\frac{2}{5}}$.
The way I see it, there are two ways to come up with an odd sum:
1. by choosing three evens and one odd
2. by choosing three odds and one even
This would seem to imply that not all of the $\binom{9}{3}=\frac{9!}{3!6!}=84$ ways of choosing the remaining three numbers, once two has been chosen, are valid. Some of these choices will leave one with two odds and two evens, which has an even sum.
When I computed the probability, I reasoned as follows
P(sum is odd) = P(three odds and 1 even) + P(three evens and one odd)
P(sum is odd) = (9/10 * 9/10 * 9/10 * 1/10) + (1/10 * 1/10 * 1/10 * 9/10)
p(sum is odd) = 738/1000
I've been wrong before about catching a mistake on Alcumus, so I think it's likely I could be wrong again. Can anyone clear this up for me?
Best Answer
You wrote that the “prime numbers are randomly selected without replacement” (emphasis mine) and therefore the prime number $2$ can be selected only once, at most. So, choosing three even primes and one odd one is not a possibility.