A driver and a passenger are in a car accident. Each of them independently has probability 0.3 of being hospitalized. When a hospitalization occurs, the loss is uniformly distributed on [0, 1]. When two hospitalizations occur, the losses are independent. Calculate the expected number of people in the car who are hospitalized, given that the total loss due to hospitalizations from the accident is less than 1. Answers Provided:
0.534

Step-by-step explanation:

p(0 losses) = 0.7² = 0.49

p(1 loss) = 2 x 0.3 x 0.7 = 0.42

p(2 losses) = 0.09

This is a conditional probability problem. If the number of people hospitalized is 0 or 1, then the total loss will be less than 1. However, if two people are hospitalized, the probability that the total loss will be less than 1 is 0.5. we need to exclude the 50% x 0.09 chance of a double loss costing more than 1. So

P(Cost < 1)

= 0.49 + 0.42 +0.045

= 0.955

P(0 losses | Cost < 1)

= P(0 losses and Cost < 1) / P(Cost < 1)

= 0.49 / 0.955 = 0.513

P(1 loss | Cost < 1)

= 0.42 / 0.955 = 0.440

P(2 losses | Cost < 1) = 0.045 / 0.955 = 0.047

Now take the expectation:

E[X] = (0)(0.513) + (1)(0.440) + (2)(0.047)

= 0.440 + 0.094

= 0.534

I get lost on how we know that with 1 hospitalization the total loss is less than 1.