Multiple choice

A bad contain $m$ white and $3$ black balls. Balls are drawn one bye one without replacement till all the black balls are drawn. The probability that this procedure for drawing balls will come to an end at the $rth$ draw is

  1. $\displaystyle \frac { \left( r-1 \right) \left( r-2 \right) }{ \left( m+1 \right) \left( m+2 \right) \left( m+3 \right) } $
  2. $\displaystyle \frac { 3\left( r-1 \right) \left( r-2 \right) }{ \left( m+1 \right) \left( m+2 \right) \left( m+3 \right) } $
  3. $\displaystyle \frac { 2\left( r-1 \right) \left( r-2 \right) }{ \left( m+1 \right) \left( m+2 \right) \left( m+3 \right) } $

  4. None of these
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B Correct answer
Explanation

The procedure ends on the rth draw when the first r - 1 draws contain exactly two black balls and the rth draw is black. Counting the possible positions of the three black balls gives 3(r - 1)(r - 2)/((m + 1)(m + 2)(m + 3)).

AI explanation

For the drawing to end on the rth draw, exactly 2 black balls must be drawn in the first (r-1) draws and the final black ball must be drawn on the rth draw, requiring the total number of balls drawn to be r and the total balls to be (m+3). The probability of selecting the last black ball at a specific position in a random sequence of the 3 black balls and (m+1) successful white ball stopping points is found using geometric probability. The resulting probability is 3(r-1)(r-2) divided by (m+1)(m+2)(m+3).

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