Both answers are correct, because there are two kinds of questions. The title only said that one person reached for two balls, but did not say whether it was at the same time. If a person catches two balls, the probability is a number, but if a person catches one ball and then reaches for the other, the probability changes again. Let me explain it separately.

(1) Suppose you reach out and touch two at the same time, that's one sixth.

We don't use probability formula for the time being, so it's easy to understand by language deduction.

Number the four balls A (red) B (red) C (yellow) D (yellow).

Then the combination of touching the ball is ab, ac, ad, bc, bd and cd. Touching two balls can only be any of these situations, because this problem is not arrangement, so ab and ba are exactly the same, and there is no order.

Obviously, two red balls appear at the same time, which can only be one of six situations, so the probability is one in six.

I hope my explanation can be understood.

(2) Catch one, catch one, and * * * catch two.

The first time I caught one, two of the four balls were red, so the percentage accounted for half.

The second catch, because the probability of the red ball being caught is the same as the remaining probability, and so is the yellow ball, which is still half, so the probability is one quarter.