Question

The number of telephone lines busy at any instant of time is a binomial variate with
probability that a line is busy is 0.1. If 10 lines are chosen at random, what is the
probability that:(i) no line is busy (ii) At least 5 lines is busy (iii) at most 3 lines are busy

Answer 1

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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−xP(X=5)=nCxpxqn−x

=10C5(0.2)5(0.8)5=0.0264=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∴P(X=2)=nCxpxqn−x

=10C2(0.2)2(0.8)8=45=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∴P(X=0)=10C0p10q0

=10C0(0.2)10(0.8)0=0.1024∗10−6=10C0(0.2)10(0.8)0=0.1024∗10−6

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