Question

An insurance company offers four different deductible levels—none, low, medium, and high—for its home owner’s policy holders and three different levels—low, medium, and high—for its automobile policyholders. The accompanying table gives proportions for the various categories of policyholders who have both types of insurance. For example, the proportion of individuals with both low homeowner’s deductible and low auto deductible is .06(6% of all such individuals).

Homeowner’s

Auto N L M H

L .04 .06 .05 .03

M .07 .10 .20 .10

H .02 .03 .15 .15

Suppose an individual having both types of policies is randomly selected.

a. What is the probability that the individual has a medium auto deductible and a high homeowner’s deductible?

b. What is the probability that the individual has a low auto deductible? A low homeowner’s deductible?

c. What is the probability that the individual is in the same category for both auto and homeowner’s deductibles?

d. Based on your answer in part (c), what is the probability that the two categories are different?

e. What is the probability that the individual has at least one low deductible level?

f. Using the answer in part (e), what is the probability that neither deductible level is low?

Short answer

a. The probability that the individual has a medium auto deductible and a high homeowner’s deductible is 0.10.

b. The probability that the individual has a low auto deductible is 0.18.

The probability that the individual has a low homeowner’s deductible is 0.19.

c. The probability that the individual is in the same category for both auto and homeowner’s deductibles is 0.41.

d. The probability that the two categories are different is 0.59.

e. The probability that the individual has at least one low deductible level is 0.31.

f. The probability that neither deductible level is low is 0.69.

Step-by-step solution

Given information

The proportions for the various categories of policyholders who have both types of insurance are provided.

Compute the probability

a.

Using the provided table,

The probability that the individual has a medium auto deductible and a high homeowner’s deductible is computed as,

\(P\left( {Auto\;M \cap Homeowner\;H} \right) = 0.10\)

Therefore, the probability that the individual has a medium auto deductible and a high homeowner’s deductible is 0.10.

b.

Using the provided table,

The probability that the individual has a low auto deductible is computed as,

\(\begin{aligned}P\left( {Auto\;L} \right) &= P\left( {Auto\;L \cap Homeowner\;N} \right) + P\left( {Auto\;L \cap Homeowner\;L} \right) + ... + P\left( {Auto\;L \cap Homeowner\;H} \right)\\ &= 0.04 + 0.06 + 0.05 + 0.03\\ &= 0.18\end{aligned}\)

Therefore, the probability that the individual has a low auto deductible is 0.18.

The probability that the individual has a low homeowner’s deductible is computed as,

\(\begin{aligned}P\left( {Homeowner\;L} \right) &= P\left( {Homeowner\;L \cap \;Auto\;L} \right) + P\left( {Homeowner\;L \cap \;Auto\;M} \right) + P\left( {Homeowner\;L \cap \;Auto\;H} \right)\\ &= 0.06 + 0.10 + 0.03\\ &= 0.19\end{aligned}\)

Therefore, the probability that the individual has a low homeowner’s deductible is 0.19.

c.

The probability that the individual is in the same category for both auto and homeowner’s deductibles is computed as,

\(\begin{aligned}P\left( {Homeowner\;L \cap \;Auto\;L} \right) + P\left( {Homeowner\;M \cap \;Auto\;M} \right) + P\left( {Homeowner\;H \cap \;Auto\;H} \right) = 0.06 + 0.20 + 0.15\\ = 0.41\end{aligned}\)

Therefore, the probability that the individual is in the same category for both auto and homeowner’s deductibles is 0.41.

d.

Using part c,

The probability that the two categories are different is computed as,

\(\begin{aligned}P\left( {two\;different\;categories} \right) &= 1 - \left( {P\left( {Homeowner\;L \cap \;Auto\;L} \right) + P\left( {Homeowner\;M \cap \;Auto\;M} \right) + P\left( {Homeowner\;H \cap \;Auto\;H} \right)} \right)\\ &= 1 - \left( {0.06 + 0.20 + 0.15} \right)\\ &= 1 - 0.41\\ &= 0.59\end{aligned}\)

Therefore, the probability that the two categories are different is 0.59.

e.

The individual has at least one low deductible level will include the events of,

\(\left( {Homeowner\;L \cap \;Auto\;L} \right)\),\(\left( {Homeowner\;L \cap \;Auto\;M} \right)\),\(\left( {Homeowner\;L \cap \;Auto\;H} \right)\),

\(\left( {Homeowner\;N \cap \;Auto\;L} \right)\),\(\left( {Homeowner\;M \cap \;Auto\;L} \right)\),\(\left( {Homeowner\;H \cap \;Auto\;L} \right)\)

The probability that the individual has at least one low deductible level is computed as,

\(\begin{aligned}P\left( {Homeowner\;L \cap \;Auto\;L} \right) + P\left( {Homeowner\;L \cap \;Auto\;M} \right) + ... + P\left( {Homeowner\;H \cap \;Auto\;L} \right) &= 0.06 + 0.10 + 0.03 + 0.04 + 0.05 + 0.03\\ &= 0.31\end{aligned}\)

Therefore, the probability that the individual has at least one low deductible level is 0.31.

f.

Using part e,

The probability that neither deductible level is low is computed as,

\(\begin{aligned}1 - \left( {P\left( {Homeowner\;L \cap \;Auto\;L} \right) + P\left( {Homeowner\;L \cap \;Auto\;M} \right) + ... + P\left( {Homeowner\;H \cap \;Auto\;L} \right)} \right) &= 1 - 0.31\\ &= 0.69\end{aligned}\)

Thus, the probability that neither deductible level is low is 0.69.