Let X denote the number of red balls in the outcome.

Possible Outcomes RR RB BR BB
X 2 1 1 0


Here, x1 = 2, x2 = 1 , x3 = 1 , x4 = 0

Now, the probability of getting `2` red balls when we draw out the balls one at a time is:

Probability of first ball being red `= 4/10`

Probability of second ball being red `= 3/9` (because there are `3` red balls left in the urn, out of a total of `9` balls left.) So:

`P(x_1)=4/10times3/9=2/15`

Likewise, for the probability of red first is `4/10` followed by black is `6/9` (because there are `6` black balls still in the urn and `9` balls all together). So:

`P(x_2)=4/10times6/9 = 4/15`

Similarly for black then red:

`P(x_3)=6/10times4/9=4/15`

Finally, for `2` black balls:

`P(x_4)=6/10times5/9=1/3`

As a check, if we have found all the probabilities, then they should add up to `1`.

`2/15 + 4/15 + 4/15 + 1/3 = 15/15 = 1`

So we have found them all.

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