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An urn contains two green, three yellow and five red balls. We draw one ball at random and put it aside. then we draw another ball until there are no more balls. Find the probability of drawing at first the two green balls, then the three yellow balls and finally the red balls

My answer:

We have two green balls so the probability of pulling two greens is: $\large\frac{1}{10} \cdot \frac{1}{9}$ for the yellow balls then we have $\large\frac{3}{8} \cdot \frac{2}{7} \cdot \frac{1}{6},$ and for the red ones we have: $1$ (since there are only $5$ balls left and there are five red balls) if we multiply all of these we get: $0.00019$

correct answer is: $0.0004$

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  • $\begingroup$ You almost have it. See this $\endgroup$ – AgentS Oct 14 '14 at 14:46
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You just made a minor mistake; the probability of pulling the two greens should be $\frac{2}{10}\cdot\frac{1}{9}$, which gives you the correct answer.

Another way to think about this problem is think of the possible order to pick the balls in. There is $1$ correct way; greens in the first two spots, yellows in the three spots after that, and the reds in the last five spots. So, the total number of ways to pick these balls is $$\binom{10}{2} \times \binom{8}{3} \binom{5}{5} = \frac{10!}{2!3!5!} = 2520$$ since you're picking two spots out of ten for the green balls, then three out of the remaining eight spots for the yellow balls, and the last five spots go to the red balls. Hence, the probability of picking the balls like that is $$\frac{1}{2520} \approx 0.0004.$$

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The probability of drawing at first one green ball is not $\frac{1}{10}$ but $\frac{2}{10}$.

So it is twice the probability you got which is roughly $0.0004$

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