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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.

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  • $\begingroup$ The numerator has two $3's$, last entry should be $2$. $\endgroup$
    – lulu
    Feb 25 '16 at 21:02
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    $\begingroup$ 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. $\endgroup$
    – lulu
    Feb 25 '16 at 21:06
  • $\begingroup$ 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). $\endgroup$ Feb 25 '16 at 21:07
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I found the answer , thanks for the commenting for the push , after just calculating Probability (first 6 balls are red) I got the desirable 1/30 which makes sence , since the only constraint is that WE DO NOT OBTAIN WHITE BEFORE THE 6TH. So I guess that was the answer to my problem , tricky words everywhere I look ...

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