A. 14,875 B 15,000 C. 15,240 D. 13,130 3. (1 point) A multi-line insurer sells 3

Question

A. 14,875 B 15,000 C. 15,240 D. 13,130 3. (1 point) A multi-line insurer sells 3 types of insurance polices: auto, home, and life. Policyholders who have two or three types of policies get a multi-line discount. Due to discount, the policyholders who have all three policies have 85% chance of renewing their policies next year. In contrast, those who have only two policies and only one policy have 70% and 50% chance of renewing their policies next year. You are given: . All polieyholders hae at keast one poliey · 51% of the policyholders have home policies · 37% of the policyholders have life policies · 15% of the policyholders have all three policies · 22% of the policyholders have both auto and home polices · 17% of the policyholders have both auto and life policies 19% of the policyholders have both honne and life policies Calculate the conditional probability that a randomly selected policyholder has only auto insurance given he renews his policy or policies next year. A. 0.13 B. 0.15 C. 0.17 D. 0.22 E. 0.27 55% are men and 45% are women, 74% of the women read

Explanation / Answer

P(R3) = 0.85, P(R2) = 0.7, P(R1) = 0.5

P(A or H or L) = P(A) + P(H) + P(L) - P(A and H) - P(A and L) + P(H and L) + P(A and B and C)

1 = P(A) + 0.51 + 0.37 - 0.22 - 0.17 - 0.19 + 0.15

P(A) = 0.55

P(A only) = 0.55 - 0.22 - 0.17 + 0.15 = 0.31
P(H only) = 0.51 - 0.22 - 0.19 + 0.15 = 0.25
P(L only) = 0.37 - 0.17 - 0.19 + 0.15 = 0.16

P(R) = P(A only)*P(R1) + P(H only)*P(R1) + P(L only)*P(R1) + P(A and H)*P(R2) + P(A and L)*P(R2) + P(H and L)*P(R2) + P(A and B and C)*P(R3)
= 0.31*0.5 + 0.25*0.5 + 0.16*0.5 + 0.22*0.7 + 0.17*0.7 + 0.19*0.7 + 0.15*0.85
= 0.8935

Required probability,
P(A only | R) = P(A only and R1)/P(R)
= 0.31*0.5/0.8935
= 0.1735

Ans: Option C (0.17)

= 0.2202

Option D: 0.22

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