From an urn, containing 6 white and 12 black balls, one takes balls randomly one by one until the second white ball appears. What is the probability that: 1) second white ball appears on the second step 2) second white ball appears on the third step 3) second white ball appears on the k-th step
My solution is the following. The probability of the second white ball is $$Pr(X=2)=\frac{6}{18}\frac{5}{17}$$
The probability of the third white ball is $$Pr(X=3)=\frac{6}{18}\frac{12}{17}\frac{5}{16}+ \frac{12}{18}\frac{6}{17}\frac{5}{16} = \frac{2⋅6⋅5⋅12}{18⋅17⋅16} $$
Therefore, the probability $$Pr(X=k)=\frac{(k-1)⋅6⋅5⋅(18-k)!⋅12!}{18!⋅(12-k+2)!}$$
Could somebody, please, check my solution, especially the third part. Thank you.
I used the hypergeometric distribution for the $(k-1)$-th draw.
The probability to draw one white ball and $k-2$ black balls in $k-1$ drawings is
$$\frac{\binom{6}{1}\cdot \binom{12}{k-2}}{\binom{18}{k-1}}=\frac{6\cdot 12!\cdot (k-1)!\cdot (19-k)!}{18!\cdot (k-2)!\cdot (14-k)!}=\frac{6\cdot 12!\cdot (k-1)\cdot (19-k)!}{18!\cdot (14-k)!}$$
At the k-th draw we have to pick a white ball. The probability is $\frac{5}{19-k}$. In total I get
$$P(X=k)=\frac{5\cdot 6\cdot 12!\cdot (k-1)\cdot (18-k)!}{18!\cdot (14-k)!}$$
This is the same result you got.