If $x_i$ where $i \in [1,6]$ are random independent variables of uniform distribution in $(0,1)$. I need to compute $P[\sum_{i=1}^{6}x_i<18]$ .
The writer says:
We have the max value when $x_i=1\, \forall i$ . Then , the max value of $\sum_{i=1}^{6}x_i=6$ hence : $P = 1$.
I was wondering where does that max value comes from?
I computed : $E[x_i]=\frac12$ and $Var(x_i) = \frac1{12}$.
Since $x_i \in U[0,1]$, $x_i$ can only take values between $0$ and $1$. In other words, the probability that $x_i$ is not between $0$ and $1$ is $0$.
$$P(x_i >1) = 0\quad P(x_i <0) = 0$$
Hence, when you add up $6$ numbers that are between $0$ and $1$, their sum can be $6$ at the most.