Question

The probability that a member of a certain class of homeowners with liability and property coverage will file a liability claim is 0.04, and the probability that a member of this class will file a property claim is 0.10. The probability that a member of this class will file a liability claim but not a property claim is 0.01. Calculate the probability that a randomly selected member of this class of homeowners will not file a claim of either type.

Answer 1

Answer:

The probability that a randomly selected member of this class of homeowners will not file a claim of either type is 0.89.

Step-by-step explanation:

Denote the events as follows:

X = liability claim will be filled

Y = property claim will be filled

The information provided is:

P (X) = 0.04

P (Y) = 0.10

P (X ∩ Y') = 0.01

The probability that a randomly selected member of the class of homeowners will not file a claim of either type will be given by:

$P[(X\cup Y)']=1-P(X\cup Y)=1-[P(X)+P(Y)-P(X\cap Y)]$

According to the law of total probability:

$P(B)=P(B\cap A)+P(B\cap A')$

Use the law of total probability to determine the value of P (X ∩ Y) as follows:

$P(X)=P(X\cap Y)+P(X\cap Y')\\\\P(X\cap Y)=P(X)-P(X\cap Y')\\\\=0.04-0.01\\\\=0.03$

The value of P (X ∩ Y) is 0.03.

Compute the value of P (X ∪ Y) as follows:

$P[(X\cup Y)']=1-P(X\cup Y)$

                   $=1-[P(X)+P(Y)-P(X\cap Y)]\\\\=1-[0.04+0.10-0.03]\\\\=1-0.11\\\\=0.89$

Thus, the probability that a randomly selected member of this class of homeowners will not file a claim of either type is 0.89.