What is the probability that $x_1+x_2+...+x_n \le n$? Given that $X_1, X_2...$ are mutually independent random variables. For each $i$ with $1\le i \le n$


*

*the variable $X_i$ is equal to either $0$ or $n+1$

*$E(X_i)$ = $1$


also.. if $X_i$ is equal to either $0$ or $n+1$, doesn't that mean that all $X_i$ need to be $0$?
Thanks
 A: $\newcommand{\+}{^{\dagger}}
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Since $\ds{X_{j} = 0\ \mbox{or}\ \pars{n + 1}\,\ \forall j = 1,2,\ldots,n}$ the solution requires $\ds{X_{j} = 0\,,\ \forall j = 1,2,\ldots,n}$. Since the probability of any $\ds{X_{J} = 0}$ is $\ds{n \over n + 1}$, the probability of $\ds{X_{j} = 0,\ \forall j}$ is given by
$\ds{\underbrace{{n \over n + 1}\ldots{n \over n + 1}}_{\ds{n\ \mbox{times}}}
=\color{#00f}{\large\pars{n \over n + 1}^{n}}}$

However, in order  to illustrate a general method we perform the following calculation which, of course, agrees with the above result:
  \begin{align}
&\color{#00f}{\large\sum_{\braces{X_{j}}}P_{1}\pars{X_{1}}\ldots P_{n}\pars{X_{n}}
\Theta\pars{n - X_{1} - \cdots - X_{n}}}
\\[3mm]&=\sum_{\braces{X_{j}}}P_{1}\pars{X_{1}}\ldots P_{n}\pars{X_{n}}
\int_{-\infty}^{\infty}
{\expo{\ic k\pars{n - X_{1} - \cdots - X_{n}}} \over k - \ic 0^{+}}
\,{\dd k \over 2\pi\ic}
\\[3mm]&=\int_{-\infty}^{\infty}{\expo{\ic kn} \over k - \ic 0^{+}}
\bracks{{n \over n + 1}\,\expo{-\ic k0}
        + {1 \over n + 1}\,\expo{-\ic k\pars{n + 1}}}^{n}
\,{\dd k \over 2\pi\ic}
\\[3mm]&=\int_{-\infty}^{\infty}
{\expo{\ic kn} \over k - \ic 0^{+}}
\bracks{n + \expo{-\ic k\pars{n + 1}} \over n + 1}^{n}\,{\dd k \over 2\pi\ic}
\\[3mm]&={1 \over \pars{n + 1}^{n}}\int_{-\infty}^{\infty}
{\expo{\ic kn} \over k - \ic 0^{+}}\sum_{j = 0}^{n}{n \choose j}n^{n - j}
\expo{-\ic jk\pars{n + 1}}\,{\dd k \over 2\pi\ic}
\\[3mm]&={1 \over \pars{n + 1}^{n}}\sum_{j = 0}^{n}{n \choose j}n^{n - j}\
\overbrace{%
\int_{-\infty}^{\infty}{\expo{\ic k\bracks{n - j\pars{n + 1}}} \over k - \ic 0^{+}}
\,{\dd k \over 2\pi\ic}}^{\ds{=\ \delta_{j0}}}\
=\color{#00f}{\large\pars{n \over n + 1}^{n}}
\end{align}
  


A: Answer:
For example, for $Y_i$ = $X_1 +X_2$ , the number of possible dintinct values could be 0, n+1, 2(n+1).  And the frequency of them is {0,0},{0,n+1},{n+1,0},{n+1,n+1}= ${2\choose0}0, {2\choose1}(n+1), {2\choose2} (2(n+1)$.
If you extend this to $Y_i = X_1 + X_2 +\cdots+ X_n$
The discrete appearances of 0, n+1, 2(n+1),... n(n+1) will follow a binomial distribution  with probability p of $(X_i = 0) = \frac{n}{n+1}$ and probability of q $(X_i = n+1) = \frac{1}{n+1}$ making the $E(X_i) = 1$. and the thus the probability
$$P(Y_i \leq n) = P(Y_i = 0)= \frac{{n\choose n}p^n.q^0}{{n\choose0}p^0.q^n + {n\choose1}p^{1}p^{n-1}+\cdots + {n\choose n}p^{n}q^{0}}$$ as 0 is the only possible value less than n.  The denominator becomes equal to one.
$$P(Y_i \leq n) = ({\frac{n}{n+1}})^{n}$$
That will be the required probability
Remark:
If $X_1, X_2,X_3,\cdots X_n$ be independent bernoulli random variables with $P(X_i = 0) = p$, Then $Y = \sum_{i = 1}^n X_i$ is a binomial random variable.
