If electricity power failures occur according to a Poisson distribution with an average of 3failures every twenty weeks.a) Calculate the probability for a given week that there will not be more than one powerfailure during a particular week?b) Calculate the probability for two consecutive weeks that there will be at most 2-powerfailure?c) Calculate the probability for a given week that there will be two or more power failureduring a particular week but less than 10 power failure.d) Calculate the probability for a year (52 weeks in a year) that there will be at most 5-power failure?

Givenon average of 3 failures for every twenty weeksfor one week =320=0.15λ=0.15 per weekx~Poisson(λ=0.15)P(X=x)=e-λ×λxx! ; x≥0a)The probability for a given week that there will not be more than one power failure during a particular week is=P(x≤1)=P(x=0)+P(x=1)=e-0.15×0.1500!+e-0.15×0.1511!=0.8607+0.1291=0.9898b)for two weeks λ=0.15×2=0.30λ=0.30 per two weeksThe probability for two consecutive weeks that there will be atmost 2-powerfailure is=P(x≤2)=P(x=0)+P(x=1)+P(x=2)=e-0.30×0.3000!+e-0.30×0.3011!+e-0.30×0.3022!=0.7408+0.2222+0.0333=0.9963c)The probability for a given week that there will be two or more power failureduring a particular week but less tha 10 power failureλ=0.15 per week=P(2≤x<10)=P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)+P(x=7)+P(x=8)+P(x=9)=e-0.15×0.1522!+e-0.15×0.1533!+e-0.15×0.1544!+e-0.15×0.1555!+e-0.15×0.1566!+e-0.15×0.1577!+e-0.15×0.1588!+e-0.15×0.1599!=0.00968+0.00048+0.00002+0+0+0+0+0=0.0102d)for 1 week there 0.15 failurefor 52 weeks =0.15×52=7.8λ=7.8 per 1year i.e., 52 weeksThe probability for a year (52 weeks in a year) that there will be atmost5 power failure isP(x≤5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)=e-7.8×7.800!+e-7.8×7.811!+e-7.8×7.822!+e-7.8×7.833!+e-7.8×7.844!+e-7.8×7.855!=0.00041+0.0032+0.01246+0.03241+0.06319+0.09858=0.2103