Question

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?

Answer 1

Explanation:

(a) Total repairs on an average in 3 years is given by:

2 repairs per year,

so , in 3 years it will be
`2* 3` = 6 repairs

Now, from Poisson distribution,

P =
`(e^(-\mu).\mu^(x))/(x!)` (1)

where,

`\mu` = no. of times an event occurs

x = 0, 1, 2, .......

Now, using eqn (1):

P(X = x) =
`(e^(-6).6^(x))/(x!)`

To calculate the probability of less than 2 repairs in the 3 years:

for
`P[X<2]`

Now,

`P[X<2] = P[X = 0] + P[X = 1]`

`P[X<2]` =
`(e^(-6).6^(0))/(0!)` +
`(e^(-6).6^(1))/(1!)`

`P[X<2]` = 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:

`P(Y \leq y)= 1 - e^(-3y)` (2)

Let Y denote the wait for next year and using eqn (2)

`P(1 \leq Y \leq 2)` =
`P(Y\leq 2) - P(Y\leq 1)`

`P(1 \leq Y \leq 2)` =
`(1 - e^(-3* 2)) + (1 - e^(-3* 1))`

`P(1 \leq Y \leq 2)` = 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