Let's create the groups and test them

1) says that More than 1/2 of the 10 employees are women.
It's clear that if all are women the probability will be more than 1/2 so we need to prove that it can be also lower.

S = {W,W,W,W,W,W,M,M,M,M}

ways to pick two people from a set of 10 C(10,2) = 10!/8!2! = 45
ways to pick two women from a set of 6 C(6,2) = 6!/4!2! = 15

Probability of picking two women = 15/45 = 1/3 which is less than 1/2

2) we need to prove that even if the probability of picking two men is less than 1/2 we have two values for the probability of picking two women one more than 1/2 and one less than 1/2

S = {W,W,W,W,W,W,W,M,M,M}

ways to pick 2 men ----> C(3,2) = 3
probability of picking two men --> 3/45 = 1/15 which is less than 1/10

ways to pick 2 women ---> C(7,2) ---> 7!/5!2! = 21
probability of picking two women = 21/45 = 7/15 < 1/2

now we need to prove that we can find a set in which the probability is more than 1/2

S = {W,W,W,W,W,W,W,W,M,M}

ways to pick 2 women ---> C(8,2) ---> 8!/6!2! = 23
probability of picking two women = 23/45 > 1/2

if you pick
S={W,W,W,W,W,W,W,W,M,M} and S{W,W,W,W,W,W,W,M,M,M}

both 1 and 2 are correct and the probabilities are one more than 1/2 and the other less than 1/2

Read Stuart's explanation of this problem if you want to see an easy and effective way to solve it:

https://www.beatthegmat.com/ds-question-t11853.html

Thank you for the link. It's really much easier.