Bernoulli distribution conditional distribution

Question

Two Independent variables have Bernoulli distribution: $X_1$ with $b(n,p)$ and $X_2$ with $b(m,p)$. How can I find conditional distribution $\mathbb P(X_1|X_1+X_2=t)$?

Answer 1

Guide (preassuming that we are dealing with binomial distribution, see my comment on the question):

$$P(X_1=k\mid X_1+X_2=t)P(X_1+X_2=t)=P(X_1=k, X_2=t-k)=P(X_1=k)P(X_2=t-k)$$where the second equality rests on independence.

Further $X_1+X_2$ will have binomial distribution with parameters $n+m$ and $p$ (this also rests on independence of $X_1$ and $X_2$ and for a proof of this see here), enabling you to find and expression for $P(X_1+X_2=t)$.

Answer 2

Hint: I assume that $X_1$ and $X_2$ are binomial distributed. We have $X_2=t-X_1$. Now you can apply the Bayes theorem. Due independency of $X_1$ and $X_2$ we get

$$P(X_1=x_1|X_1+X_2=t)=\frac{P(X_1=x_1)\cdot P(X_2=t-x_1)}{P(X_1+X_2=t)}$$

where $X_1+X_2\sim Bin(n+m,p)$

Answer 3

Assuming you meant Binomial random variables and that they are independents, you can always write $X_1=\sum_{i=1}^n Y_i$ and $X_2=\sum_{i=n+1}^{n+m} Y_i$ where the random variables are independent Bernoulli random variable with parameter $p$. Once you have that then $X_1+X_2=\sum_{i=1}^{n+m} Y_i$. The event $X_1+X_2=t$ just indicate that $t$ of the $Y_i$ are equal to $1$, conditioned on that the event $X_1=s$ is then an indication that $s$ of the first $n$ $Y_i$ are one. Observe that this implies that $t-s$ of the last $m$ $Y_i$ are one, hence the probability is just the number of way of choosing $s$ Bernoulli out of the $n$ first and $t-s$ Bernoulli out of the last $m$, we divide this by the amount of ways of having $t$ of the Bernoulli equal to $1$, hence \begin{align*} \mathbb P(X_1=s|X_1+X_2=t) = \frac{{n\choose s}{m\choose t-s}}{{m+n\choose t}} \end{align*}