Respuesta :
The question is incomplete. The complete question is :
The Jackson family is undecided about whether or not to buy a new car. If the probability is .9 that they will buy one, and if the probability is .3 that they will buy a Ford, and if the probability is .4 that they will purchase a car getting more than 20 miles per gallon, what is the probability that they will buy either a car getting more than 20 miles per gallon or a Ford, if all Fords get more than 20 miles per gallon?
Solution :
Given that :
The probability of buying a new car, [tex]P(NC) = 0.9[/tex]
Probability of buying Ford = 0.3
That is, if Jackson family buy a car that is a ford car, [tex]$P(F) = 0.9 \times 0.3$[/tex]
= 0.27
The probability for getting more than 20 miles per gallon = 0.4
That is if Jackson family buy a car that have more than 20 miles per gallon mileage, [tex]$P(20) = 0. 9 \times0.4=0.36$[/tex]
The conditions
All of the car have more than 20 miles per gallon mileage.
It means that buying a ford car is subset of getting more than 20 miles per gallon.
[tex]$P(20 \text{ miles per gallon}\ \cap \ \text{Ford})=P(F)$[/tex]
Therefore, the probability of buying a car either getting more than 20 miles per gallon or ford = [tex]$P(20 \text{ miles per gallon}\ \cup \ \text{Ford})$[/tex]
Therefore,
[tex]$P(20 \text{ miles per gallon}\ \cup \ \text{Ford})=P(20) + P(F) - P(20 \text{ miles per gallon }\cap \ \text{Ford})$[/tex]
[tex]$P(20 \text{ miles per gallon}\ \cup \ \text{Ford})=P(20) + P(F) - P(F)$[/tex]
[tex]$P(20 \text{ miles per gallon}\ \cup \ \text{Ford})=P(20) $[/tex]
= 0.36
Thus the probability that Jackson family is buying a car either getting more than 20 miles per gallon or ford is 0.36
