Note: I have edited the question to add more context to it. Please provide me with feedback.
One unbiased die is thrown 10 times. For $1 \leq i \leq 6$, let $X_i$ denote the number of times $i$ appears.
Find the conditional expectation of $X_1$ given that $X_6 = 5$.
My approach to solve this problem is as follows:
The formula for computing the expectation is:
$E(X)= \sum_{i=1}^n x_{i}P(X=x_i)$ where $X$ is the random variable and $x_i$'s are the values that the random variable takes. $P(X=x_i)$ is the probability of the random variable taking the value $x_i$.
Now, coming back to the question, it is clear that the random variable $X_1$ can take any value from 0 to 5, with 0 and 5 included. This is because we are already given that the value of the random variable $X_6$ is 5, and the total number of throws is 10.
Now, the next task is to compute the probability distribution of $X_1$ given that $X_6$ = 5.
I first try to compute the joint probability distribution for $X_1$ and $X_6$.
This has already been done in a past question, the link to which is attached as follows:
Regarding probability distribution
We know that $P(A|B)= \frac{P(A \cap B)}{P(B)}$.
So by making use of the knowledge gained from the attached question and the definition of conditional probability, we get $P(X_1=i|X_6=5)= \frac{{10 \choose i}(1/6)^i {10-i \choose 5} (1/6)^5 (4/6)^{5-i}}{{10 \choose 5}(1/6)^5 (5/6)^5}$ when $0 \leq i \leq 5$ and 0 otherwise.
In order to compute the expectation, we then make use of the formula:
$E(X_1|X_6=5)= \sum_{i=0}^5 i P(X_1=i|X_6=5)$
I want to understand if there is an easier way to compute the same.
Best Answer
$$10=E[X_1+X_2+...+X_6|X_6]=5E[X_1|X_6]+X_6$$ $$\quad E[X_1|X_6]=\frac{10-X_6}{5},\quad E[X_1|X_6=5]=\frac{10-5}{5}=1$$