Question

The probability that at any moment one telephone line out of 10 will be busy 0.2. (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?

Answer 1

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)=nCxpxqn−x

=10C5(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

∴P(X=2)=nCxpxqn−x

=10C2(0.2)2(0.8)8=45

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

∴P(X=0)=10C0p10q0

=10C0(0.2)10(0.8)0=0.1024∗10−6

∴ The probability that all the lines are busy is 0.1024 *10−6

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