Question

Suppose that in a year, 15% of all light vehicles (SUVs, pickups, passenger cars, and minivans) sold in the United States were pickups and 55% were passenger cars. Moreover, a randomly chosen vehicle sold that year was twice as likely to be an SUV as a minivan. Find the associated probability distribution.

Answer 1

The probability distribution is:

P(pickup)=0.15

P(passenger cars)=0.55

P(minivan)=0.10

P(SUV)=0.20

Explanation:

We know that 15% of all light vehicules were pickups. Then P(pickup)=0.15.

We also know that 55% were passenger cars, so P(passenger cars)=0.55.

The left probability is lef for SUV and minivans.

It is a total of P(SUV or Minivans)=1-0.15-0.55=0.30.

We also know that a randomly chosen vehicle sold that year was twice as likely to be an SUV as a minivan. This is equivalent to P(SUV)=2*P(minivan).

Then we have:

P(SUV or Minivans)=P(SUV)+P(minivan)=0.30

2*P(minivan)+P(minivan)=0.30

3*P(minivan)=0.30

P(minivan)=0.10

P(SUV)=2*P(minivan)=2*0.10=0.20