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From $15$ yellow, $7$ white and $9$ red roses were made bouquets of $3$ roses randomly-selected. What is the probability in one bouquet all roses to be red?

I translated the problem from my language, so I am not sure if I made it clear, but I have some questions.

Let $А=\{$in one bouquet all roses are red$\}$. All possible outcomes for $A$ are $n_A=C_{5+7+9}^3=C_{21}^3=1330$. The favorable outcomes are $m_A=C_9^3=84$. Now $P(A)=\dfrac{m_A}{n_A}=\dfrac{84}{1330}=\dfrac{6}{95}$. Do you understand the problem in this way?

What if we have $21$ roses ($9$ red and $12$ not red). From them were made $7$ bouquets of $3$ roses. What is the probability that one of these $7$ bouquets will be entirely red?

Math Student
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  • Assume you mean exactly one. I think you may need to count the number of red roses in different bouquets as $(3, X_1, X_2, \ldots, X_6)$. Since you require exactly one, you can enumerate the remaining from $(1, 1, 1, 1, 1, 1)$, $(2, 1, 1, 1, 1, 0)$ etc. – BGM Apr 04 '20 at 11:31

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I found an answer to the more general problem extended stars-and-bars problem(where the upper limit of the variable is bounded)

Number of bouquet permutations is given by number of solutions to $\sum_{i=1}^7b_i=9$ where $b_i\in\{0,1,2,3\}$ which is the coefficient of $x^9$ in the polynomial $(1+x+x^2+x^3)^7$ and the number of permutations with one 3-red bouquet is 7 times the number of solutions to $\sum_{i=1}^6b_i=6$ where $b_i\in\{0,1,2\}$, because only one of the 7 bouquets/numbers has 3 red roses and there are 6 left to distribute among the remaining 6. $p=\frac{7\times141}{1918}$

Ymh
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