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?
Best Answer
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. $$\dfrac{\dbinom 62\sum\limits_{\imath=0}^2\dbinom 4\imath 24^{\imath}}{26^6}$$