Probability of sixth ball to be white 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
 A: 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
$$\frac{\binom{5}{3}\cdot\binom{6}{2}}{\binom{12}{5}}=\frac{25}{132}.$$
A: I would make two remarks here,
1) The balls may not be identical. For each type of extraction like $000111001111$ there are exactly $7!5!$ labeled cases and this factor is the same for good configuration, or bad, or all configurations, such that the ratio in the probability is not affected. We may securely count labeled configurations.
2) There is no need to draw the balls one at a time, but we can grab them more at once, following a $6+5+1$ grabbing rule.
These having been said, I will write two e.g.f.'s that describe the good configurations among all.
$\frac{good} {all} = \frac {E_2(X)E_3(Y)}{E_5(X+Y)} \frac {X}{X+Y} \frac {E_4(X)E_2(Y)} {E_6(X+Y)}  $
Above is the species presentation. The e.g.f's are
$good = \frac {x^2y^3}{2!3!} {x} \frac {x^4y^2}{4!2!}  $
$all ={ (x+y)^5 \over 5!} (x+y) { (x+y)^6 \over 6!} $
Now we compare the coefficients of $x^7y^5\over 7!5! $ in the two e.g.f.'s above and we get:
$\frac {1050}{5544} = \frac{25}{132}$
