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Could you confirm that the answer for (a) is (6C3)*(5C1)*(4C1)/(15C5) instead of (6C3)*(5C1)*(4C1)/(15C4) in the book? Thanks,
The question was supposed to say "$2$ red, $1$ green, and $1$ yellow", since we are picking $4$ balls total. Thus the answer would be $$\dfrac{\binom{6}{2} \binom{5}{1}\binom{4}{1}}{\binom{15}{4}}.$$
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