Question

10% of the downtown residents of a town in the Plains live in mobile homes. Based on data from past tornadoes, if a tornado goes through downtown and a person lives in a mobile home, there is a 0.3 probability the person will die as a result of the tornado. However, if a tornado goes through downtown and a person lives somewhere other than a mobile home, there is only a 0.02 probability the person will die as a result of the tornado. What is the probability a person will not live in a mobile home if they do not die as a result of the tornado

Answer 1

The probability a person will not live in a mobile home if they do not die as a result of the tornado is 0.9265.

Explanation:

Denote the events as follows:

X = a person lives in a mobile home.

Y = a person dies as a result of tornado.

Given:

`P(X)=0.10\\P(Y|X) = 0.30\\P(Y|X^(c))=0.02\\`

Compute the value of
`P(Y^(c)|X^(c))` as follows:

`P(Y^(c)|X^(c))=1-P(Y|X^(c))\\=1-0.02\\=0.98`

Compute the probability that a person dies as a result of tornado as follows:

`P(Y)=P(Y|X)P(X)+P(Y|X^(c))P(X^(c))\\=(0.30*0.10)+(0.02*(1-0.10))\\=0.03+0.018\\=0.048`

Then the probability that a person does not dies as a result of tornado is:

`P(Y^(c))=1-P(Y)\\=1-0.048\\=0.952`

Compute the value of
`P(X^(c)|Y^(c))` as follows:

`P(X^(c)|Y^(c))=(P(Y^(c)|X^(c))P(X^(c)))/(P(Y^(c)))=(0.98*(1-0.10))/(0.952)=0.9265`

Thus, the probability a person will not live in a mobile home if they do not die as a result of the tornado is 0.9265.