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

a.

Using the provided table,

The probability that the individual has a medium auto deductible and a high homeowner鈥檚 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鈥檚 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鈥檚 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鈥檚 deductible is 0.19.

c.

The probability that the individual is in the same category for both auto and homeowner鈥檚 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鈥檚 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.