Answer to Question #149655 in Statistics and Probability for mark

Question

Sport utility vehicles (SUVs) vans, picks up are generally considered to be more prone to roll over than cars. In 1997, 24.0% of all highway fatalities involved rollovers: 15.8% of all fatalities in 1997 involved SUVs vans and pickups given that fatality involved a rollover. Given that a rollover was not involved 5.6% of all fatalities involved SUVs vans and pickups.
Consider the following definitions: A= fatality involved an SUV van or pickup B=fatality involved an rollover
a. If a highway fatality is selected at random, what is the probability that the fatality involved an SUV a van or pickup?
b.Use Bayes' theorem to find the probability that a fatality involved a rollover given that the fatality involved an SUV a van or pickup.

Answer 1

We have that A= fatality involved an SUV van or pickup and B=fatality involved a rollover

Then 24.0% of all highway fatalities involved rollovers: $P(B) = 24\% = 0.24$

15.8% of all fatalities in 1997 involved SUVs vans and pickups given that fatality involved a rollover: $P(A | B) = 15.8\% = 0.158$

Given that a rollover was not involved 5.6% of all fatalities involved SUVs vans and pickups: $P(A | \bar B) = 5.6\% = 0.056$

a) the probability that the fatality involved an SUV a van or pickup:

$P(A) = P(A | B)P(B)+P(A | \bar B)P(\bar B) = 0.158\cdot0.24+0.056\cdot0.76=0.0805$

where $P(\bar B)=1-P(B)=1-0.24=0.76$

b) the probability that a fatality involved a rollover given that the fatality involved an SUV a van or pickup using Bayes' theorem:

$P(B|A)=\frac{P(A|B)P(B)}{P(A)}=\frac{P(A|B)P(B)}{P(A | B)P(B)+P(A | \bar B)P(\bar B)}=\frac{0.158\cdot0.24}{0.0805}=0.4711$

Answer:

a) the probability that the fatality involved an SUV a van or pickup is 0.0805

b) the probability that a fatality involved a rollover given that the fatality involved an SUV a van or pickup is 0.4711