# Probability of Sum of 4 of 1st 10 Primes Being Odd

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?

• There is only one even prime. – lulu Sep 23 '19 at 12:51
• I think the key words you may have overlooked is "selected without replacement" – a_student Sep 23 '19 at 12:53
• Your computation is hard to follow. You appear to be assuming replacement, though the problem is quite clear that there is no replacement. Also you seem to be specifying the order in which the odd and even primes are chosen, counter to your intent. Also your denominator is incorrect. – lulu Sep 23 '19 at 12:53
• @lulu I see the error that I made in assuming there was replacement. But I don't think the order matters in my calculation, does it? Multiplication is commutative, after all – David J. Sep 23 '19 at 12:58
• Just think about it. $OOOE, \,EOOO, OEOO, OOEO$ are four separate events. True, they each have the same probability, but you have to consider all four of them! – lulu Sep 23 '19 at 13:10

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.
Sum of four odd numbers is always even. The sum of an even and an odd number is always odd. The sum of four prime numbers out of the first ten prime numbers can only be odd if it contains $$2$$ as one of the four prime numbers.