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Let's say I have G green balls, Y yellow balls and R red balls. I'm interested in probability distribution of selecting particular amount of green and yellow balls.

For example, if G=10, Y=5 and R=20, what is the probability of drawing 2 green and 3 yellow when drawing 10 in total.

For two colors I would use hypergeometric distribution, but I'm a bit stuck in incorporating the two different "success" possibilities.

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2 Answers 2

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(Assuming without replacement.) The number of ways of choosing a "good" combination of the balls is $$n_{\text{good}}:=\binom{G}{2}\binom{Y}{3}\binom{R}{10-2-3}$$ and the total number ways of choosing $10$ balls is $$n_{\text{total}}:=\binom{G+Y+R}{10}.$$ So the probability is $n_{\text{good}}/n_{\text{total}}$.

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We solve the numerical problem, in the expectation that generalization will be clear.

So we have $10$ green, $5$ yellow, and $20$ red, and want to find the probability of getting $2$ green and $3$ yellow, when drawing $10$ balls, presumably without replacement.

Think of the balls as distinct (they have student numbers). Then there are $\binom{35}{10}$ ways to choose $10$ balls, all equally likely.

There are $\binom{10}{2}$ ways of getting $2$ green, and for each of these ways $\binom{5}{3}$ ways to get $3$ yellow. For each of these ways, there are $\binom{20}{5}$ ways to choose the $5$ red needed.

This gives a total of $\binom{10}{2}\binom{5}{3}\binom{20}{5}$. For the probability, divide by $\binom{35}{10}$.

Remark: If drawing is done with replacement, we get the situation that you find familiar. The probability of getting $2$ green, $3$ yellow, and $5$ red is then $$\frac{10!}{2!4!4!}\left(\frac{10}{35}\right)^2\left(\frac{5}{35}\right)^3\left(\frac{20}{35}\right)^5.$$

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