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

Answer:

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.

Step-by-step 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:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})\\=(0.30\times0.10)+(0.02\times(1-0.10))\\=0.03+0.018\\=0.048$

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:

$P(X^{c}|Y^{c})=\frac{P(Y^{c}|X^{c})P(X^{c})}{P(Y^{c})}=\frac{0.98\times(1-0.10)}{0.952}=0.9265$

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.