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There are three red balls, three white balls and three blue balls in an urn. We will randomly pick three of these balls. What's the probability that we get one ball of each color?

Solution: Total numbers of ways to pick three balls = $9 \choose 3$

Selecting one red ball = $3 \choose 1$

Selecting one white ball = $3 \choose 1$

Selecting one blue ball = $3 \choose 1$

Number of ways to picking one ball of each color = $\dbinom{3}{1}\dbinom{3}{1}\dbinom{3}{1} $

Probability of picking one ball of each color = $\frac{{3 \choose 1} {3 \choose 1}{3 \choose 1}}{9 \choose 3}$

Did I solve the problem correctly?

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  • $\begingroup$ This is okay. What doubts are you having? $\endgroup$ – Graham Kemp Sep 30 '16 at 6:54
  • $\begingroup$ I was using intuition to solve it, and I wasn't sure if I did it correctly. $\endgroup$ – Quazi Irfan Sep 30 '16 at 7:00
  • $\begingroup$ Yeah...absolutely correct!!! $\endgroup$ – tatan Sep 30 '16 at 7:01
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Yes.   You solved the problem nicely.

If in doubt, expand it out:

$$\dfrac{\binom 3 1 \binom 3 1\binom 3 1}{\binom 9 3} = \dfrac{3!\cdot 3\cdot 3\cdot 3}{\quad\;\; 9\cdot 8\cdot 7}$$

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  • $\begingroup$ Can you explain how do you get the numerator in simpler terms? (right hand side) $\endgroup$ – Quazi Irfan Sep 30 '16 at 7:03
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    $\begingroup$ @iamcreasy $\dfrac{\binom 3 1\binom 3 1\binom 3 1}{\binom 9 3} = \dfrac{3\cdot 3\cdot 3}{\frac {9!}{3!~6!}} = \dfrac{3\cdot 3\cdot 3}{\frac{9\cdot 8\cdot 7}{3!}}$. $\endgroup$ – Graham Kemp Sep 30 '16 at 7:36

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