Question

In studying the insurance preferences of homeowners the following conclusions are made: A homeowner is three times as likely to purchase additional jewelry coverage as additional electronics coverage The purchase of one of these coverages is independent of the purchase of the other The probability a homeowner purchases both coverages is 0.2 Find the probability that a homeowner purchases exactly one coverage.

Answer 1

Let J be the event that a homeowner purchases additional jewel coverage and E be the event that a homeowner purchases additional electronics coverage.

P(J) = 3P(E)

If two events A and B are independent, P(A)xP(B) = P(A & B)

Here, Since E and J are independent,

P(J) x P(E) = P(E & J) = 0.2

3P(E) x P(E) = 0.2

P(E) = 0.258

P(J) = 0.774

P(Homeowner purchases exactly one coverage) = P(E) + P(J) - 2P(E & J)

= 0.258 + 0.774 - 2x0.2

= 0.632