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  • Assume that on an average,one telephone out of 10 is busy .Ten telephone numbers are randomly selected and called. Find the probability that three of them will be busy.

Assume that on an average,one telephone out of 10 is busy .Ten telephone numbers are randomly selected and called. Find the probability that three of them will be busy.

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Let A be the event "telephone is dialled and it is busy",then
p=P(A)=frac110,so, q=frac910,and, n=7
Thus,we have binomial distribution with p=frac110,q=frac910,and, n=7
Required probability=P(3)=7C3p3q4
=frac7 imes6 imes51 imes2 imes3 imes(frac110)^3 imes(frac910)^4Rightarrow frac35 imes9^410^7

Posted by

Deependra Verma

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