<s>[INST] Conditional probability and the binomial distribution 
Let an unbiased die be cast at random seven independent times. Compute the conditional probability that each side appears at least once given that side 1 appears exactly twice.

We have $P[(\text{Each side appears} \geq 1) \mid (\text{Side 1 appears exactly  exactly twice})] = \dfrac{P[(\text{Each side appears} \geq 1) \cap (\text{Side 1 appears exactly twice})]}{P[\text{Side 1 appears exactly twice}]}.$
For the numerator, if side 1 appears exactly twice, then the other 5 sides must appear exactly once. The probability of side 1 appearing exactly twice and the other 5 sides appearing exactly once is $(\frac{1}{6})^2(\frac{1}{6})^5$. There are $\binom{7}{2}$ ways of having side 1 appear exactly twice, and there are $5!$ ways the other 5 sides can appear exactly once.
Hence the numerator is  $\binom{7}{2}\cdot5!\cdot(\frac{1}{6})^2(\frac{1}{6})^5.$
For the denominator, the 1 appearing exactly once has the probability $(\frac{1}{6})^2(\frac{5}{6})^5$. Considering that the 1's can be rolled in $\binom{7}{2}$ ways, the denominator is $\binom{7}{2}(\frac{1}{6})^2(\frac{5}{6})^5$.
When I divide the numerator by the denominator, I get $120\cdot\frac{1}{5^7} = \frac{24}{5^6},$ but the author has $\frac{24}{625} = \frac{24}{5^4}$, so somewhere I have two too many factors of 5.
 [/INST] Looks like you just made a little mistake near the end of your calculation: instead of $120 \cdot \frac1{5^7}$, it should be $120 \cdot \frac1{5^5}$.
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