# Probability of mutually independent random variables [duplicate]

I have been doing an exam review and this is one of the problems I could not solve.

Let $X_1, X_2, ..., X_n$ be a sequence of mutually independent random variables.

For each $i$ with $1 \leq i \leq n$,

• the variable $X$, is equal to either $0$ or $n+1$
• $E(X_i) = 1$

Determine $Pr(X_1+ X_2 + ... + X_n \leq n)$

## marked as duplicate by Did, Avitus, Umberto P., Najib Idrissi, user63181 Apr 2 '14 at 18:52

• Again? This is an epidemic... – Did Apr 2 '14 at 17:01

Hint: (i) Find $\Pr(X_i=n+1)$; (ii) Our sum is $\le n$ if and only if all the $X_i$ are equal to $0$.

If $X$ could be $0$ or $n+1$ you can call $p$ to the probability of being $0$ and obviously $1-p$ is the probability of being $n+1$

How the expectation is equal to $1$:

$$E(X)=p\cdot 0 + (1-p)(n+1)=1$$ $$p=\frac{n}{n+1}$$

For one side: $$\sum_{i=0}^n X_i \leq n \Rightarrow X_i\leq n , \forall i=1..n\Rightarrow X_i=0, \forall i=1..n$$ For other: $$X_i=0, \forall i=1..n \Rightarrow\sum_{i=0}^n X_i=0 \leq n$$

Then: $$\sum_{i=0}^n X_i \leq n \iff X_i=0, \forall i=1..n$$

Then:

$$P\left(\sum_{i=0}^n X_i \leq n\right)=P(X_i=0, \forall i=1..n)=p^n=\left(\frac{n}{n+1}\right)^n$$

The last condition is accomplished via the independence of the variables.