Question

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.

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.

Answer 1

Solution

Let X denote the number of lines in use at a specified time.

Calculate the probability of each of the following

(a) At most three lines are in use.

\( \implies P(X\leq 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3) \)

                             \( =0.10+0.15+0.20+0.25 \)

                             \( =0.7 \)

Therefore. \( P(X\leq 3)=0.7 \)

(b) Fewer than three lines are in use.

\( \implies P(X<3)=P(X\leq 3)-P(X=3) \)

                             \( =0.7-0.25=0.45 \)

Therefore. \( P(X<3)=0.45 \)

(c) At least three lines are in use.

\( \implies P(X\geq 3)=1-P(X<3)=1-0.45=0.55 \)

Therefore. \( P(X\geq 3)=0.55 \)

(d) Between two and five lines, inclusive,are in use.

\( \implies P(2\leq X\leq 5)=P(2)+.....+P(X=5) \)

                                    \( =0.2+0.25+0.2+0.06=0.71 \)

Therefore.\( \hspace{2mm}P(2\leq X\leq 5)=0.71 \)

(e) Between two and four lines, inclusive,are not in use.

X : number of lines in use .

The number of lines are not in use is  6-X

\( \implies P(2\leq 6-X\leq 4)=P(2\leq X\leq 4)=P(2)+.....+P(4) \)

                                           \( =0.20+0.25+0.20=0.65 \)

Therefore. \( P(2\leq X\leq 4)=0.65 \)

(f) At least four lines are not in use.

\( \implies P(6-X\geq 4)\iff P(X\leq2)=P(X=0)+P(X=1)+P(X=2) \)

                                                                  \( =0.10+0.15+0.20=0.45 \)

Therefore. \( P(X\leq 2)=0.45 \)

Therefore. 

a). \( P(X\leq 3)=0.7 \)

b). \( P(X\geq 3)=0.55 \)

c). \( P(2\leq X\leq 5)=0.71 \)

d). \( P(2\leq X\leq 4)=0.65 \)

e). \( P(X\leq 2)=0.45 \)