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

Answer:

0.632

Step-by-step explanation:

Given that a homeowner is three times as likely to purchase additional jewelry coverage as additional electronics coverage

If probability of purchasing additional electronics coverage = p, then prob of purchasing jewelry coverage = 3p

The two events are independent hence joint probability is product of these two.

i.e. P(both) = $p(3p) =3p^2$

This is given as 0.2

$3p^2 =0.2\\p = 0.258$

the probability that a homeowner purchases exactly one coverage.

$=P(AUB)-P(A \bigcap B)$

= Prob he purchases I + prob he purchases II-2(prob he purchases both)

$= 0.258+3(0.258)-2(0.2)\\=0.632$