Create a probability distribution of socks

[F.B]

A drawer contains four red socks and two blue socks. Three socks are drawn from the drawer without replacement.
a)Create a probability distribution in which the random variable represents the number of red socks.
b)Determine the expected number of red socks if three are drawn from the drawer without replacement.

Ok i need help with a).
This is what i did.

C(6,3) = 40 which is the total.

Now for since there are only two blue socks you have to draw atleast 1 red for 0 red socks.

C(4,1) x C(2,2)

I don't know if i did that right because it says without replacement. So how do i take into account that it is done without replacement.

 

[HallsofIvy]

A drawer contains four red socks and two blue socks. Three socks are drawn from the drawer without replacement.
a)Create a probability distribution in which the random variable represents the number of red socks.
b)Determine the expected number of red socks if three are drawn from the drawer without replacement.
Ok i need help with a).
This is what i did.
C(6,3) = 40 which is the total.
Now for since there are only two blue socks you have to draw atleast 1 red for 0 red socks.
C(4,1) x C(2,2)
I don't know if i did that right because it says without replacement. So how do i take into account that it is done without replacement.

Obviously, since there are only two blue socks, you can't draw 3 blue sock so you can't draw 0 red socks:
P(0)= 0.

How can you draw exactly one red sock? One way is to draw "Red", "Blue", "Blue" in that order. First draw one red sock: the probability of that is 4/6= 2/3. Now, ASSUMING you drew a red sock first, there are 3 red socks and 2 blue socks. The probability that the next sock you draw is blue is 2/5. Given that that happens, there are 3 red socks and 1 blue sock left. The probability of drawing a blue sock is now 1/4. That is: the probability of drawing "Red", "Blue", "Blue" in that order is (2/3)(2/5)(1/4)= 1/15.
It should be easy to see that the probability of "Blue", "Red", "Blue" and of "Blue", "Blue", "Red" are exactly the same and so P(1)= 3(1/15)= 1/5.
Notice that that 3 is ₃C₁.

Now you need to find P(2) and P(3). You can draw exactly two red socks as "Red", "Red", "Blue" or "Red", "Blue", "Red", or "Blue", "Red", "Red".
Find the probability of one of those as above and multiply by 3= ₃C₂.

You can draw exactly three red socks as "Red", "Red", "Red". Find the probability of that, P(3), as above.