Question

Let us suppose that some article studied the probability of death due to burn injuries. The identified risk factors in this study are age greater than 60 years, burn injury in more than 40% of body-surface area, and presence of inhalation injury. It is estimated that the probability of death is 0.003, 0.03, 0.33, or 0.88, if the injured person has zero, one, two, or three risk factors, respectively. Suppose that three people are injured in a fire and treated independently. Among these three people, two people have one risk factor and one person has three risk factors. Let the random variable X denote number of deaths in this fire. Determine the cumulative distribution function for the random variable. Round your answers to five decimal places (e.g. 98.76543).

Answer 1

Answer:

The cumulative distribution is:

P(X=0)=0.11291

P(X≤1)=0.94788

P(X≤2)=0.99921

P(X≤3)=1.00000

Step-by-step explanation:

First, we can write the sample space for X

$X=[0,1,2,3]$

For X=0:

- There is one combination possible (no one dies).

- The probability is

$P(0)=(1-P_1)^2*(1-P_3)=(0.97)^2*0.12=0.11291$

For X=1:

- There are 3 combinations possible, one for each person (one death). One combination has different probability than the other (when the person with three risk factor dies).

- The probability is:

$P(1)=(1-P_1)^2P_3+2(1-P_1)*P_1*(1-P_3)\\\\P(1)=0.97^2*0.88+2*0.97*0.03*0.12\\\\P(1)=0.82799+0.00698=0.83498$

For X=2:

- There are 3 combinations, with different probabilities (two persons die).

- The probability is:

$P(2)=P_1^2(1-P_3)+2(1-P_1)P_1P_3\\\\P(2)=0.03^2*0.12+2*0.97*0.03*0.88\\\\P(2)=0.00011+0.05121=0.05132$

For X=3:

- There is only one combination (the three persons die)

- The probability is:

$P(3)=P_1^2P_3=0.03^2*0.88=0.00079$

The cumulative distribution is:

P(X=0)=0.11291

P(X≤1)=0.11291+0.83498=0.94788

P(X≤2)=0.94788+0.05132=0.99921

P(X≤3)=0.99921+0.00079=1.00000