Looks good to me. In your numerator you have the number of potential committees that contains $1$ woman (and two men), added to the number of potential committees that contain $3$ women (and no men). In the denominator you have the total number of potential committees. Since each potential committee is equally likely, this gives the probability that there are either exactly one, or exactly three women in the committee.
The actual calculation is correct too, which always helps.
As a teacher, I have two things I would've liked to see. One is a sentence or two (like my paragraph above, but you don't have to be that thorough) explaining where all the numbers come from. The other is that I would've preferred a $=$ rather than $\Rightarrow$, since that's what you have: two fractions with the same value.
Exactly how important these are is something only your teacher can tell you, though. I am not the one correcting your tests, hand-ins and exams.