# Computing the probability that at least four telephone lines are not in use.

A mail-order computer business has six telephone lines. Let $X$ denote the number of lines in use at a specified time. Suppose the pmf of $X$ is as given in the accompanying table: $p(0)=.10$, $p(1)=.15$, $p(2)=.2$, $p(3)=.25$, $p(4)=.2$, $p(5)=.06$, $p(6)=.04$

Calculate the probability of $\{\text{at least four lines are not in use}\}$.

I interpreted the problem as find $P(4 \leq X \leq 6)$ but I'm not getting the correct answer. Can someone help me see what I'm doing wrong?

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We want $P(0 \le X\le 2)$. Remember it said not in use.
To help you understand it, think of this $$P(0) = P(\text{ 6 lines not in use} )\\ P(1) = P(\text{ 5 Lines not in use })\\ P(2) = P(\text{ 4 Lines not in use })$$ Then $$P(\text{ 6 Lines not in use })+P(\text{ 5 Lines not in use })+P(\text{ 4 Lines not in use })= \\ P(0)+P(1)+P(2)=0.1+0.15+0.2=0.45$$