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A store sells 26 flavors or ice cream(A-Z). We choose six flavors at random(repeats allowed). Find the probability that there are 2 of flavor A and at least 2 of flavor B.

My approach: Of the six we choose, 2 for sure need to be flavor A. Then of the four remaining, at least two of them to be flavor B.

$$\sum_{i=2}^{4}{6 \choose 2}{4 \choose i}\left(\frac{1}{26}\right)^2\left(\frac{1}{26}\right)^i\left(\frac{24}{26}\right)^{4-i}$$

Is this correct?

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  • $\begingroup$ There is a lot of sense in your answer, but you should explain where the terms come from to help people assess the logic. Presumably the $6 \choose 2$ is choosing which two scoops are $A$, but you should say that. At least when I was in school, decades ago, that was expected. MathJax hint: if you want parentheses to grow to encompass whatever is inside, make them \left( and \right), so \left(\frac 1{26}\right) gives $\left(\frac 1{26}\right)$. It looks a little nicer. It is not so important for small fractions. $\endgroup$ – Ross Millikan Jun 23 '20 at 2:29
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    $\begingroup$ On simplifying your answer, we get $$\frac{\displaystyle \sum_{i=2}^4 {6\choose 2}{4\choose i}(24)^{4-i}}{(26)^6}$$ which is absolutely correct. $\endgroup$ – SarGe Jun 23 '20 at 2:37
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Is this correct?

Yes it is. $$\dfrac{\dbinom 62\sum\limits_{\imath=2}^4\dbinom 4\imath 24^{4-\imath}}{26^6}$$

You could also specify that you need 2 from 6 scoops to be of flavour A and up to 2 from 4 remaining scoops to be selected from the 24 other flavours. $$\dfrac{\dbinom 62\sum\limits_{\imath=0}^2\dbinom 4\imath 24^{\imath}}{26^6}$$

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