Two red, one green, one yellow candy

Question

A bowl holds 5 red candies, 4 green candies, and 3 yellow candies, 12 in total. You grab 4 candies at random. What is the probability of grabbing exactly 2 red, 1 green, and 1 yellow?

Accepted Answer

With three colors, the same combination recipe extends: choose the required count from each color's pile separately, multiply those counts, then divide by all ways to grab the handful. Recall \binom{n}{k} is the number of ways to choose k from n. Total ways to grab 4 candies from 12: \binom{12}{4}=495. Favorable ways are the product of choosing within each color: \binom{5}{2}	imes\binom{4}{1}	imes\binom{3}{1}=10	imes 4	imes 3=120. Divide: \frac{120}{495}=\frac{8}{33}\approx 0.242. Takeaway. For a target mix across several colors, choose each color's share from its own pile, multiply the c…

Two red, one green, one yellow candy

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