Answer to Question #152478 in Statistics and Probability for Yosef

Question

An urn contains 8 red balls. 3 yellow, 9 white, 6 balls

are randomly selected with replacement

What is the probability 2 are red l is yellow, and 3 are

white?

Answer 1

We can take $2$ of $8$ red balls by $\binom{8}{2}$ ways, $1$ of $3$ yellow balls by $\binom{3}{1}$ ways, and $3$ of $9$ white balls by $\binom{9}{3}$ ways, so we can take $2$ red, $1$ yellow and $3$ white balls by $\binom{8}{2}\binom{3}{1}\binom{9}{3}$ ways.

In general we can take $6$ of $20$ balls by $\binom{20}{6}$ ways.

So our probability is $\frac{\binom{8}{2}\binom{3}{1}\binom{9}{3}}{\binom{20}{6}}=\frac{\frac{8!}{2!6!}\cdot\frac{3!}{2!1!}\cdot\frac{9!}{3!6!}}{\frac{20!}{6!14!}}=\frac{28\cdot 3\cdot 84}{38760}=\frac{294}{1615}$

Answer: $\frac{294}{1615}$