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a) probability of choosing 3 of same color?

I am aware that the probability is 3*C(5,3)/(15,3).

But how would you do it fraction wise? The first pick would be 15/15 because its any of the 15 objects. Then it would be 10/15 and 5/15. So fraction wise, to find the answer, shouldn't be (10/15)*(10/15) and that will yield 50/225? It doesn't give me the same answer as 3*C(5,3)/(15,3). What am I doing wrong? Thanks

b) also how can you solve 3 of different color fraction wise, not choosing wise. Thanks

Roy Kesserwani
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1 Answers1

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Imagine choosing the balls one at a time. Whatever was the first ball chosen, the probability the next ball matches it in colour is $\frac{4}{14}$. And given that the first $2$ balls match in colour, the probability that the third ball matches the first $2$ is $\frac{3}{13}$. Thus our required probability is $\frac{4}{14}\cdot\frac{3}{13}$.

For $3$ different colours, whatever was the first colour chosen, the probability that the next ball is of a new colour is $\frac{10}{14}$. And given the second ball is of a new colour, the probability the third ball is new is $\frac{5}{13}$, for a probability of $\frac{10}{14}\cdot\frac{5}{13}$.

Remark: The fact that there are simple "fractionwise" solutions is due to the symmetries in the problem: all colours occur the same number of times. If the numbers of occurrences are different, the "choosing" version is simpler to handle than the "fractionwise" approach.

André Nicolas
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