Question

There are 50 telephone lines in an exchange. The probability that any of them will be busy is 0.1. The probability that all the lines are busy is

Answer 1

Solution

We have:

P(telephone line is busy ) = 0.01

Q(telephone lines are not busy) = 1 - 0.01 = 0.99

N =5

So it is problem of binomial distribution

1) The probability for all lines are busy (which is equivalent to no lines are free) = P(X=0)=

5

C

0

p

5

q

0

=0.01

5

=10

−10

2) The probability for more than 3 lines are busy = P(X=0)+P(X=1)=

5

C

0

p

5

q

0

+

5

C

1

p

4

q

1

=(.01)

5

+5(.01)

4

(0.99

1

)

=10

−10

+495×10

−10

=496×10

−10

=4.96×10

−10

Answer 2

Given:

50 telephone lines in an exchange

probability that any of them will be busy is 0.1

we have

$\text { Let } x \text { be the Poisson variable, "no of busy lines in the exchange". }$

$\mathrm{P}(x=r)=\frac{e^{-m} m^{r}}{r !}, r=0,1,2, so \ on$

$\text { Here } n=50, p=0.1 \quad\\\\ \therefore \quad m=n p=50 \times 0.1=5$

$\mathrm{P}(x=r)=\frac{e^{-5}(5)^{r}}{r !}, r=0,1,2, \ldots \ldots, 50$

$\therefore \quad \mathrm{P}(\text { all lines are busy })\\\\=\mathrm{P}(x=50)=\frac{e^{-5}(5)^{50}}{50 !}$

Hence the probability that all the lines are busy is  $\frac{e^{-5}(5)^{50}}{50 !}$