A mail-order company 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
\begin{array}{r|cccccccc}
x & 0 & 1 & 2 & 3 & 4 & 5 & 6\\\hline
p(x)&0.12&0.15&0.20&0.25&0.18&0.09&0.01
\end{array}
Calculate the probability of each of the following events.
(a) {at most three lines are in use}
$\boxed{.72 \qquad}\,\checkmark$
(b) {fewer than three lines are in use}
$\boxed{.47\qquad}\,\checkmark$
(c) {at least three lines are in use}
$\boxed{.53\qquad}\,\checkmark$
(d) {between two and five lines, inclusive, are in use}
$\boxed{.72\qquad}\,\checkmark$
(e) {between two and four lines, inclusive, are not in use}
$\boxed{0\quad\!\qquad}\,\mathsf X$
(f) {at least four lines are not in use}
$\boxed{0\quad\!\qquad}\,\mathsf X$
Hello, I am trying to calculate probabilities based on a given distribution which is indicated by the probability distribution table in the picture above.
As you can see, there are 2 problems that I got wrong that I need help on.
For the 5th problem I thought the answer would be .37 because the probability that between 2 and 4 (inclusive) would not be in use would be the added probabilities of 0, 1, 5, and 6, but that was wrong too.
Thanks.
Best Answer
If there are $k$ lines not in use, then it is the same as $6-k$ lines being in use.
If between $2$ and $4$ lines, inclusive, are not in use, then that is the same as between $2$ and $4$ lines, inclusive, being in use. ($2$ lines not being in use is the same as $4$ lines being in use, etc.)
If at least $4$ lines are not in use, then this is the same as at most $2$ lines being in use.