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.

 

[Architeuthis Dux]

a) To create a probability distribution of socks, we need to first determine the possible outcomes of drawing three socks from the drawer without replacement. These outcomes can be represented as follows:

1) All three socks are red
2) Two socks are red and one is blue
3) One sock is red and two are blue
4) All three socks are blue

To calculate the probability of each outcome, we can use the formula P(A) = (number of ways A can occur) / (total number of possible outcomes).

1) P(All three socks are red) = C(4,3) / C(6,3) = 4/20 = 1/5
2) P(Two socks are red and one is blue) = C(4,2) x C(2,1) / C(6,3) = 6/20 = 3/10
3) P(One sock is red and two are blue) = C(4,1) x C(2,2) / C(6,3) = 4/20 = 1/5
4) P(All three socks are blue) = C(2,3) / C(6,3) = 0

Therefore, the probability distribution for the number of red socks is as follows:

Number of red socks | Probability
0 | 0
1 | 1/5
2 | 3/10
3 | 1/5

b) To determine the expected number of red socks, we can use the formula E(X) = Σ (x * P(x)), where X is the random variable and P(x) is the probability of getting x red socks.

E(X) = (0 * 0) + (1 * 1/5) + (2 * 3/10) + (3 * 1/5) = 0 + 1/5 + 6/10 + 3/5 = 1 + 9/10 = 1.9

Therefore, the expected number of red socks is 1.9 when three socks are drawn without replacement from the drawer.