Lesson: Determining Independence

P(both selections are women) = P(1st selection is a woman AND 2nd selection is a woman)
= (4/7)(3/6)
= 12/42
= 2/7

P(1st toss is heads AND 2nd toss is heads)
= P(1st toss is heads) x P(2nd toss is heads)
= 1/2 x 1/2
= 1/4

Your solution is great, EXCEPT for the numerator (it's not 3C2).

P(both selections are women) = (number of ways to select 2 women from the FOUR women)/(number of ways to select 2 people from the 7 people)
= (4C2)/(7C2)
= 6/21
= 2/7

That's true!

Here's the question: "Suppose that Adam rolls a fair six-sided die and a fair four-sided die simultaneously. Let A be the event that the six-sided die is an even number, and let B be the event that the four-sided die is an odd number."

Notice that the probability of event B (getting an odd number on the 4-sided die) is NOT affected by whether or not event A occurs.
Regardless of what happens with the 4-sided die, P(B) is ALWAYS 1/2

Compare this to the following question:
A bag contains 5 red balls and 1 yellow ball.
Joe randomly select one ball (without replacement), and then another ball.
Let event A be selecting the yellow ball on the first draw.
Let event B be selecting the yellow ball on the second draw.
What is the probability of event B?
Well, it DEPENDS.
If event A (getting a yellow ball on the first draw) occurs, then P(event B) = 0, since the yellow ball has already been taken.
If event A does NOT occur, then P(event B) = 1/5, since there are now 5 balls remaining, and 1 of them is yellow.
As you can see, the probability of event B occurring DEPENDS on whether event A occurred.