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.

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

what is the original number? you forgot to add it.

you didn't give me the choices, so i will just give you the graph so you can look at your choices and choose the right one.