2) On average, a house has 2 repairs per year. a) What is the probability that in the next 3 years, there will be less than 2 repairs? b) I had 3 repairs this year. What is the probability that I have to wait between 1 and 2 more years for my next repair?

Step-by-step explanation:(a) Total repairs on an average in 3 years is given by:2 repairs per year,so , in 3 years it will be [tex]2\times 3[/tex] = 6 repairsNow, from Poisson distribution, P = [tex]\frac{e^{-\mu}.\mu^{x}}{x!}[/tex]          (1)where, [tex]\mu[/tex] = no. of times an event occursx = 0, 1, 2, .......Now, using eqn (1):P(X = x) = [tex]\frac{e^{-6}.6^{x}}{x!}[/tex]To calculate the probability of less than 2 repairs in the 3 years: for [tex]P[X<2][/tex]Now,[tex]P[X<2] = P[X = 0] + P[X = 1][/tex] [tex]P[X<2][/tex] = [tex]\frac{e^{-6}.6^{0}}{0!}[/tex] + [tex]\frac{e^{-6}.6^{1}}{1!}[/tex] [tex]P[X<2][/tex] = 0.01735(b)  To calculate the probability for 1 and 2 more years for next repair, if current repairs are 3:The distribution function can be given by:[tex]P(Y \leq y)= 1 - e^{-3y}[/tex]                 (2)Let Y denote the wait for next year and using eqn (2)[tex]P(1 \leq Y \leq 2)[/tex] = [tex]P(Y\leq 2) - P(Y\leq 1)[/tex][tex]P(1 \leq Y \leq 2)[/tex] = [tex](1 - e^{-3\times 2}) + (1 - e^{-3\times 1})[/tex][tex]P(1 \leq Y \leq 2)[/tex] = 0.04731Answer:(a) The probability for less than 2 repairs in next 3 years is 0.01735(b) The probability to wait between 1 and 2 more years for next repair is 0.04731