Question

If the chance that any of $5$ telephone lines is busy at any instance is $0.01$. What is the probability that all the lines are busy? What is the probability that more than $3$ lines are busy?

Answer 1

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 = $P(X=0)+P(X=1)={}^5{C}_0{p}^5{q}^0+{}^5{C}_1{p}^4{q}^1=(.01{)}^5+5({.01)}^4\left({0.99}^1\right)={10}^{-10}+495\times {10}^{-10}=496\times {10}^{-10}=4.96\times {10}^{-10}$