Probability / Calculation
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All the fractions can be reduced, I didn't bother as the reduction is trivial.
P(X=2) = (2/6)*(1/5) = 2/30
Explanation: (R,R) is the only possible way for this to happen,
The first R is drawn with prob. 2/6, the 2nd with 1/5.
P(X=3) = ((2/6)(4/5) + (4/6)(2/5)) * (1/4) = 4/30
Explanation: There are two ways this can happen: (R,*,R) or
(*,R,R), where * is any non-red ball. Notice how it must
end with R, and that last R has probability 1/4 since there
will be 4 balls left when that one is chosen.
P(X=4) = ((2/6)(4/5)(3/4) + (4/6)(2/5)(3/4) + (4/6)(3/5)(2/6)) * (1/3) = 6/30
Explanation/Notes: we could have assumed (R,*,*,R) and for the (R,*,*)
portion you have 3 places to put the R: 3*(2/6)(4/5)(3/4)
= 3*(6/30) (then *(1/3) for the 2nd R on the 4th selection
gives you 6/30).
By now you should see the pattern...
P(X=5) = 8/30
P(X=6) = 10/30
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Let's re-write the PDF:
P(X=2) = 2/30
P(X=3) = 4/30
P(X=4) = 6/30
P(X=5) = 8/30
P(X=6) = 10/30
and (2+4+6+8+10)/30 = 30/30 = 1, as expected
There may be more elegant solutions, but this 'brute force' method also works.
