How to calculate conditional probability in Bernoulli trial?

Question

Let a fair die is thrown $10$ times. Let $X_i$ denotes the number of times the digit $i$ appears. What would be the probability $P(X_2=2|X_1=2)$ ?

By formula $P(X_2=2|X_1=2)=\frac{P(X_1=2~\cap~ X_2=2)}{P(X_1=2)}$.

$P(X_1=2)$ means the probability that the digit $1$ appears twice, which is $\binom{10}{2}\left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^8.$

But since the events $X_1=2$ and $X_2=2$ are independent, we have $P(X_1=2 \cap X_2=2)=P(X_1=2)P(X_2=2)$=$\binom{10}{2}\left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^8 \cdot \binom{10}{2}\left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^8.$

Thus, $P(X_2=2|X_1=2)=\frac{P(X_1=2~\cap~ X_2=2)}{P(X_1=2)}=\binom{10}{2}\left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^8$.

But the answer is given $\binom{8}{2}\left(\frac{1}{5}\right)^2 \left(\frac{4}{5}\right)^6$. How is this possible ?

Answer 1

You are told that you got two 1's in 10 rolls of the die, so in the remaining 8 rolls you have no 1's and two 2's. Assuming that 5 outcomes $\{2..6\}$ are equally likely (because the die is fair) leads directly to the answer given in your book $$P(X_2=2|X_1=2)=\binom{8}{2}\left(\frac{1}{5}\right)^2 \left(\frac{4}{5}\right)^6$$

Answer 2

$X_1$ and $X_2$ are not independent. You can think this if $X_1=10$ that will affect $X_2$. In this case $X_2$ has to be zero. $\mathbb{P}(\{X_1=2\}\cap \{X_2=2\})=\frac{10!}{2!2!6!}(\frac{1}{6})^2(\frac{1}{6})^2(\frac{4}{6})^6$.

Answer 3

A shortcut may be to say that conditioned on $X_1=2$, looking at the other dice you must have $X_2 \sim \textrm{Bin}(8,\frac15)$ and so $$\mathbb P(X_2=2 \mid X_1=2)= {8 \choose 2}\frac{4^6}{5^8}=0.29360128.$$

This is the same as using Karambir.kd's result in $\frac{\mathbb P(\{X_1=2\} \cap \{X_2=2\})}{\mathbb P(X_1=2)} = \frac{\frac{10!}{2!2!6!}(\frac{1}{6})^2(\frac{1}{6})^2(\frac{4}{6})^6}{\frac{10!}{2!8!}(\frac{1}{6})^2(\frac{4}{6})^8}$