Probability ,balls from urn question Quite confused, I want to know if im on the right track, and if so, how to explain it,
The question:
In an urn we have 7 black balls and 5 white balls, we draw a ball at **random ,and put it back in the urn time after time,what is the probability that on the 10th time we get a black ball is on the 20th draw?**
Intuitively im thinking that the probability is 7/12, because it doesnt matter if take a black ball first or 10th...but I really hope someone can direct me in the right direction.
 A: When answering such questions, you need to distinguish the event given, i.e. that has already occurred, and the event which followed, whose probability you have to find.
The event that has already occurred here would be - out of the first $19$ balls picked, $9$ were black. Lets call this event A.
Now event B would be that on the $20^{th}$ draw, we got a black ball.
Since after every time you pick a ball, it is put back in the set, the two events are independent.
P(B/A) = P(A)P(B)
P(A)= $\binom{19}{9}(\frac{7}{12})^{9}(\frac{5}{12})^{10}$
P(B)=$7/12$
Plug in the values to get your answer.
A: From @lulu's comment, $$P(\,exactly\,\,9\,\,black\,balls\,in\,first\,\,19\,\,draws\, ) = \binom{19}{9}p^{9}(1-p)^{10}$$
$$P(\,black\, ball\,) = p = \frac{7}{12}$$
First, you must pick exactly 9 black balls out of the 19 draws, then you pick the last one on the 20th draw.  Since these two events independent, we just multiply the two:
$$P(\,10th\,black\, ball\, on\, 20th\, draw\, ) = \binom{19}{9}p^{9}(1-p)^{10} * p \approx 0.0665$$
Which is what I get when I plug into wolframalpha.  Sorry if my formatting looks weird, I am new to MSE and still learning how to format my answers.  I hope this helps!  
