Question

Due date is today after 2 hous

Assume that three business owners each own an identical storage building valued at $120,000.

Assume that there is a 1% chance in any year each building will be destroyed by a peril (fire), and that a loss to any of the buildings is an independent event.

Assuming that:

(first case ) Fire events are independent of each other

(second case) Each building faces the same types of risks and environmental conditions.

Given that, solve the below questions for each case:

1. Use the compound probability of independent events law to find the sample space of the probability of each possible event

2. Find the annual expected loss for each of the insureds.

3. Find the risk of incurrence of losses (Std dev)

Answer 1

1.Let B1 be the outcome that building 1 is burnt, B2 be the outcome that building 2 is burnt, B3 be the outcome that building 3 is burnt. Similarly, B1^c be the event that 1 is not burnt and so on.

An event would consist of all 3 buildings whether they are burnt of not. Ex- (B1, B2, B3) means all 3 buildings are burnt. (B1, B2^c, B3^c) represents 1 is burnt but not 2 and 3.

Hence, the sample space of events is S = {(B1,B2,B3) , (B1^c,B2,B3), (B1,B2^c,B3), (B1,B2,B3^c), (B1^c,B2^c,B3), (B1,B2^c,B3^c) , (B1^c,B2,B3^c), (B1^c,B2^c,B3^c)}

P(B1,B2,B3) = 0.01*0.01*0.01 = 0.000001

P(B1^c,B2,B3) = 0.99*0.01*0.01 = 0.000099

P(B1,B2^c,B3) =  0.01*0.99*0.01 = 0.000099

P(B1,B2,B3^c) = 0.000099

P(B1^c,B2^c,B3) = 0.99*0.99*0.01 = 0.009801

P(B1,B2^c,B3^c) = 0.009801

P(B1^c,B2,B3^c) = 0.009801

P(B1^c,B2^c,B3^c) = 0.99*0.99*0.99 = 0.970299

2.Annual expected loss for each of the insureds = P(B)*Loss + P(B^c)*No-loss = 0.01*120000 + 0.99*0

= 1200

3.Variance for each building = (120000 - 1200)² * 0.01 + (0 - 1200)² * 0.99 = 142560000