Respuesta :
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 repairs
Now, from Poisson distribution,
P = [tex]\frac{e^{-\mu}.\mu^{x}}{x!}[/tex] (1)
where,
[tex]\mu[/tex] = no. of times an event occurs
x = 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] = 0.04731
Answer:
(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
