The problem is:

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.

$$\begin{array}{c|c|c} x & \text{0} & \text{1} & \text{2} & \text{3} & \text{4} & \text{5} & \text{6}\\ \hline \\p(x) & .10 & .15& .20 & .25 & .20 & .06 & .04 \end{array}$$

Calculate the probability of each of the following events.

a. {at most three lines are in use}

b. {fewer than three lines are in use}

c. {at least three lines are in use}

d. {between two and five lines, inclusive, are in use}

e. {between two and four lines, inclusive, are not in use}

f. {at least four lines are not in use}

The only parts I was unable to do were parts e) and f).

For part f), I removed the negation of the statement, hoping that that might make the original statement more clear. So, if we say that we want the probability of the at least being in use, that would be the probability of $4$ in use, $5$ in use, or $6$ in use. But we don't care to know that probability, we want to know the opposite: we want to know the probability of less than 4 lines in use. This process didn't get me the right answer, however.

Could someone help me with those two parts?


Isn't f the same as b? For a, you are counting if $0,1,2,$ or $3$ lines are in use, so just add $0.10+0.15+0.20+0.25=0.70$ The rest are similar.

  • $\begingroup$ Should you be counting $3$ in the calculation of "fewer than 3 lines are in use"? Shouldn't it be 0,1,2 ? $\endgroup$ – Inquest Feb 16 '13 at 0:07
  • $\begingroup$ @Inquest: a calls for at most 3, so 3 is included. b is the one that wants fewer than 3-I didn't do that one but it looks the same as f to me. $\endgroup$ – Ross Millikan Feb 16 '13 at 0:08
  • $\begingroup$ Oops. I misread your answer. I thought you meant F is same as B and B is same as A. My bad. Apologies. $\endgroup$ – Inquest Feb 16 '13 at 0:11
  • $\begingroup$ @Inquest: no problem $\endgroup$ – Ross Millikan Feb 16 '13 at 0:12
  • (E) : Probability that 1 5 or 6 are in use. Just sum them.
  • (F) : Probability that at least four lines are not in use. This one is tricky one, I think of it this way. What is the probability that out of 6 people you know: at least 4 are not busy > at least 4 are free > 4 are free or 5 or 6 are free > 0,1 or 2 friends are busy > 0,1 and 2 lines are busy. This is the same as (B).
  • $\begingroup$ Are you using ">" to denote that statements are equivalent? $\endgroup$ – Mack Feb 16 '13 at 0:24
  • $\begingroup$ Yes. I am using it to mean "verbally implies". Is it sloppy? $\endgroup$ – Inquest Feb 16 '13 at 0:37
  • $\begingroup$ No, it's fine, now that I know what it means. Thank you for responding. $\endgroup$ – Mack Feb 16 '13 at 1:25

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