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An urn has $5$ white and $10$ black balls. A die is rolled, and that many balls is drawn from the >urn. What is the probability that all the balls drawn are white?

My thinking is that each die roll $(d)$ has probability $\frac{1}{6}$. The probability to get only white balls from your draw is $$\frac{C(5, d)}{C(15, d)}$$. Then you add up all the probabilities. I come up with $\frac{5}{66}$, but have no way of knowing if I'm right. Is this the way to go about it? I tried multiplying the probabilities tied to each die roll and came up with an extremely small chance, so that seems wrong to me.

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What kind of game is rolling a die ? :-D – Luc M Feb 5 '14 at 15:03
up vote 5 down vote accepted

Your answer is right. This probability is in fact $$\frac{1}{6}\cdot\frac{{5 \choose 1}}{{15 \choose 1}}+\dots+\frac{1}{6}\cdot\frac{{5 \choose 5}}{{15 \choose 5}} = \frac{1}{6}\cdot\left(\frac{1}{3} + \frac{2}{21} + \frac{2}{91} + \frac{1}{273}+ \frac{1}{3003}\right) = \frac{1}{6}\cdot\frac{5}{11} = \frac{5}{66}\text.$$

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Your idea looks right to me. As the die roll gets large, the chance of all white balls gets very small. For the first two, we have $\frac 16\cdot \frac 5{15}+\frac 16\cdot \frac5{15}\cdot \frac 4{14}=\frac 1{18}\cdot \frac {18}{14}=\frac 1{14}=\frac 5{70}$, which is not much less than $\frac 5{66}$

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