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There is a bin with 1 green ball, 1 red ball, 1 blue ball, and 1 yellow ball.

We draw 4 balls with replacement. What is the probability of drawing exactly 2 of a given color? Order doesn't matter, as long as we have exactly 2 balls drawn of a given color.

My attempt was as follows:

There are $\binom 4 2$ ways to select two "spots" in the 4 output spots. For each of those ways, the probability of having exactly two is $\frac 1 4 \cdot \frac 1 4 \cdot \frac 3 4 \cdot \frac 3 4$. Hence the final probability is just $\binom 4 2 \cdot \frac 1 4 \cdot \frac 1 4 \cdot \frac 3 4 \cdot \frac 3 4$.

Is this right?

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    $\begingroup$ Yes, this looks good. You could also see this as an instance of the binomial distribution with success probability $\frac 14$. $\endgroup$
    – lulu
    Sep 14 at 20:58
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Suppose the 'given color' is Red. The probability of Red on any one draw is $1/4.$ Let $X$ be the number of times Red is drawn. Then $X \sim \mathsf{Binom}(n=4, p=1/4).$ We seek $P(X=2) = 0.2109.$

dbinom(2, 4, .25)
[1] 0.2109375

According to this interpretation RRYG, RBRY, and RYRY all meet the condition.

However, if the condition is that exactly two balls of any color are drawn, then RRYG, GBBY both count. That seems to be a harder problem. This may be a possible interpretation of the title, as it currently stands.

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  • $\begingroup$ Sorry, the second list should have been something like RRYG and GBBY. Fixing the typo. $\endgroup$
    – BruceET
    Sep 15 at 7:26

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