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I have $10$ identical black balls, $5$ identical red balls, $2$ identical white balls in my box. What is the probability of choosing three black balls from this box?

My Attempt: We have to first consider each ball is different and the points in the sample space is ${17} \choose 3$.

We can choose three black balls in ${10} \choose 3$ ways. So the probability is $\frac{{10} \choose 3}{{17} \choose 3}$

Can anyone please check my attempt ?

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    $\begingroup$ The question should mention "in three draws", then it is correct. $\endgroup$ Mar 3, 2021 at 7:14
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    $\begingroup$ @trueblueanil This solution is for 1 draw of three balls, at the same time, or three consecutive draws without replacement. But that's how I interpret it. $\endgroup$ Mar 3, 2021 at 8:07
  • $\begingroup$ @HennoBrandsma: Yes, I agree, the solution is right for such a question. $\endgroup$ Mar 3, 2021 at 8:55

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The answer is correct, if you draw the three balls without putting them back. To be even clearer, you can also write the solution this way: $$\frac{\binom{10}{3}\binom{5}{0}\binom{2}{0}}{\binom{17}{3}} = \frac{\binom{10}{3}}{\binom{17}{3}}$$

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