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We don't have an answer key for this test questions and I just wanted to confirm my answers.

We have a bag with 6 balls, 3 blue, 2 green, and 1 red.

If we select 3 at random, what is the chance we select only 2 of 3 are blue balls?

So I have there are a total of $\binom{6}{3} = 20$ different ways to select 3 balls. There is $\binom{3}{2} * \binom{2}{1} = 6$ ways of selecting 2 blue and 1 green, and $\binom{3}{2} * \binom{1}{1} = 3$ of selecting 2 blue and 1 red.

So we have a probability of $\frac{6 + 3}{20} = \frac{9}{20} $ chance of selecting only 2 blue balls.

Also, what is the chance of selecting only 1 of each color?

$\binom{3}{1} * \binom{2}{1} * \binom{1}{1} = 3*2*1 = 6$ different ways of selecting one of every color. So we have a probability of $\frac{6}{20} $ of getting 1 of each color.

Is this correct?

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  • $\begingroup$ Keep in mind this solution holds iff all the balls are distinct, e.g. enumerated $\endgroup$ – Alex Dec 3 '16 at 13:49
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Yes, that is okay.   $9/20$ and $6/20$ are the answers for the reasons you gave.


The probability of selecting exactly $2$ from $3$ blue ball (and $1$ from $3$ non-blue) when selecting $3$ from all $6$ balls is more simply $\dbinom 3 2\dbinom 31\big/\dbinom 6 3$.

But yes, $\dbinom 3 2\Big(\dbinom 2 1 +\dbinom 1 1\Big)\big/\dbinom 6 2$ is also valid.   It counts the same thing.


The probability of selecting $1$ ball from each group of sizes $3,2,1$ when selecting $3$ from all $6$ balls, is indeed $\dbinom 3 1\dbinom 2 1\dbinom 1 1\big/\dbinom 63$

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  • $\begingroup$ Thank you for the alternate way of thinking for the first problem. $\endgroup$ – Kaska Nov 30 '16 at 2:50

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