Question

A bag contains m white and 3 black balls. Balls are drawn one by 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
A.$\dfrac{{(r - 1)(r - 2)}}{{(m + 1)\left( {m + 2} \right)(m + 3)}}$
B.$\dfrac{{3(r - 1)(r - 2)}}{{(m + 1)\left( {m + 2} \right)(m + 3)}}$
C.$\dfrac{{2(r - 1)(r - 2)}}{{(m + 1)\left( {m + 2} \right)(m + 3)}}$
D.None of these

Answer 1

Hint: Probability for any random experiment is given by:
\[{\text{Probability of an event}} = \dfrac{{{\text{Number of occurence of event A in S}}}}{{{\text{Total number of cases in S}}}} = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}\]
In this question, we need to determine the probability of the process of drawing the balls without replacement until all the 3 black balls come out of the bag.


Complete step by step solution:
For the procedure of drawing balls one by one without replacement to come to an end at the rth draw, 2 black balls out of the 3 must be drawn in previous (r-1) draws.
Event or Favourable space is \[^3{C_2}{ \times ^m}{C_{r - 3}}\]
Sample space or Total ways is \[^{m + 3}{C_{r - 1}}\]
Probability of the event is evaluated as:
$
  P = \dfrac{{^3{C_2}{ \times ^m}{C_{r - 3}}\;\;\;\;}}{{^{m + 3}{C_{r - 1}}}} \times \dfrac{1}{{m + 3 - \left( {r - 1} \right)}} \\
   = \dfrac{{3!}}{{2!1!}} \times \dfrac{{m!}}{{\left( {m - \left( {r - 3} \right)} \right)!\left( {r - 3} \right)!}} \times \dfrac{{(r - 1)!\left( {m + 3 - \left( {r - 1} \right)} \right)!}}{{\left( {m + 3} \right)!}} \times \dfrac{1}{{m - r + 4}} \\
   = 3 \times \dfrac{{m!}}{{\left( {m - r + 3} \right)!(r - 3)!}} \times \dfrac{{\left( {m - r + 4} \right)!(r - 1)!}}{{(m + 3)!}} \times \dfrac{1}{{m - r + 4}} \\
   = 3 \times \dfrac{{m!}}{{(m + 1)(m + 2)(m + 3)m!}} \times \dfrac{{\left( {m - r + 4} \right)\left( {m - r + 3} \right)!}}{{\left( {m - r + 3} \right)!}} \times \dfrac{{(r - 1)(r - 2)(r - 3)!}}{{(r - 3)!}} \times \dfrac{1}{{m - r + 4}} \\
   = \dfrac{{3(r - 1)(r - 2)}}{{(m + 1)\left( {m + 2} \right)(m + 3)}} \\
 $
Option (B) is correct


Note: Combinations are a way to calculate the total outcomes of an event where an order of the outcomes does not matter. To calculate combinations, we use the formula $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
A shortcut method to find the value of \[n{C_r}\]is given below as:
$n{C_r} = \dfrac{{n(n - 1)(n - 2).......(n - r + 1)}}{{r!}}$
The above trick can be easily proved by doing simple calculations in the original equat