# What is the probability that the first white ball is seen after the $6$th draw?

Hey guys a really easy question that I solved but the solution says otherwise so I need to check if the solution is wrong (hope so).

An urn contains $3$ white balls, $7$ red balls. Balls are drawn one by one without replacement. What is the probability that the first white ball is seen after the 6th draw?

First does this mean the $6$th is white or, after the sixth draw, meaning the seventh?

Anyhow, it means the preceding balls are red so after doing the calculation I keep on getting $\frac 1{40}$. But in the solution (MCQ) it says $\frac 1{30}$ But how can I make a mistake with such a simple question : MY answer was : $$\frac {7\times6\times5\times4\times3\times3}{10\times9\times8\times7\times6\times5}$$ Took the assumption that there are $5$ reds before, so the sixth is white.

• The numerator has two $3's$, last entry should be $2$.
– lulu
Feb 25 '16 at 21:02
• The problem does not specify where the first white ball is seen, only that it is "after the sixth draw". Thus, it is asking for the probability that the first $6$ draws are all red.
– lulu
Feb 25 '16 at 21:06
• It means the first six balls are red, and thus that the first white ball appears some time after that. (which will be either the seventh or eighth place). Feb 25 '16 at 21:07