A box contains 7 identical white balls and 5 identical black balls. They are to be drawn randomly one at a time without replacement until the box is empty. Find the probability that the 6th ball drawn is white, while before that exactly 3 black balls are drawn.
Source : Principle and techniques in combinatorics by Chen Chuan Chong , Ch 1, question 24
My Approach
Let '1' denote a white ball and '0' denote a black ball.
Consider following sequence: 1 1 0 0 0 1 (i.e., first two balls drawn are white then next three balls are black and finally a white ball)
The probability of this event is: $\displaystyle \frac 7{12} \cdot \frac 6{11} \cdot \frac 5{10} \cdot \frac 49 \cdot \frac 38 \cdot \frac 27 \cdot = \frac 1{132}$
Now, there are 10 binary sequences in which 6th digit is 1 and before them three digits are 0 hence, probability should be $\displaystyle \frac {10}{132}$
Correct Answer: $\displaystyle \frac{25}{132}$
Please indicate my mistake and if possible please give a complete solution.
Thank You
Best Answer
Henry pointed out your error. This is another apprach.
There are $\binom{12}{5}$ ways to extract in order the $12$ balls. The number of extractions where there are $3$ black balls in the first $5$ and the remaining $2$ black balls in the last $6$ is $\binom{5}{3}\cdot\binom{6}{2}$. So the probability should be $$\frac{\binom{5}{3}\cdot\binom{6}{2}}{\binom{12}{5}}=\frac{25}{132}.$$