i. What is the probability that 5 lines are busy?

ii. Find the expected number of busy lines and also find the probability of this number

iii. What is the probability that all lines are busy?

Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV

Marks: 6M

Year: Dec 2015

We have:

P(telephone line is busy ) = 0.2

Q(telephone lines are not busy) = 1 -- 0.2 = 0.8

N =10

So it is problem of binomial distribution

1) The probability for exact 5 lines to be busy :

$$P(X=5) = n_{C_x}p^xq^{n-x}$$ $$ = {10}_{C_5} {(0.2)}^5{(0.8)}^5 = 0.0264$$

2) Expected number of busy lines : E(X) = np =0.2 *10 =2

Probability of 2 lines to be busy

$$\mathrm{\therefore } P(X=2) = n_{C_x}p^xq^{n-x}$$ $$ = {10}_{C_2} {(0.2)}^2{(0.8)}^8 = 45$$

3) P(X) all lines are busy is equivalent to no line is free.

$$\mathrm{\therefore } P(X=0) = {10}_{C_0}p^{10}q^0$$
$$ = {10}_{C_0} {(0.2)}^{10}{(0.8)}^0 = 0.1024 * {10}^{-6}$$

$\mathrm{\therefore }$ The probability that all the lines are busy is 0.1024 *${10}^{-6}$