A bag contains 10 white and 3 black balls. Balls are drawn one by one without replacement till all the black balls are drawn. Then find the probability that this procedure for drawing the balls will come to an end at the rth draw.

Explanation

Procedure of drawing th balls has to end at the rth draw. So, exactly two black balls are drawn in first (r-1) draws and third black ball is drawn in rth draw.
Let event A be" In first (r-1) draws, two black balls are drawn" and evet B be "rth draw is black ball"
We have to find `P(AnnB).` Now, `P(AnnB)=P(A)P(B//A)`
P(A)=P(2black and (r-3) white balls are drawn from (W+3)balls)
`=(""^(3)C_(2).^(""w)C_(r-3))/(""^(W+3)C_(r-1))`
`=((3!)/(2!!!).(W!)/((r-3)!(W-r+3!)))/(((W+3)!)/((r-1)!(w-r+4)!))`
`=(3.(r-1)(r-2)(W-r+4))/((W+3)(W+2)(W+1))`
At the end of (r-1)th draw, we woluld be left with 1 black and `(W-r+3)` white balls.
`therefore P(B//A)=P` (drawing the black ball at the rth draw)
`(1)/((W-r+4))`
Therefore. `P(AnnB)=P(A)P(B//A)`
`=(3(r-1)(r-2)(W-r+4))/((W+3)(W+2)(W+1)(W-r+4))`
`=(3(r-1)(r-2))/((W+3)(W+2)(W+1))`