Question

The probability that a new car of a particular make will have a brake failure is given in the table.

Car Make Brake Failure
in New Car
A 0.0065%
B 0.0037%
C 0.0108%
D 0.0029%
E 0.0145%
Total 0.0048%


If a given car has a brake failure, what is the probability that it is make D

Answer 1

Ok so are you looking for the answer in a fraction form

Answer 2

Do you know anything about the probability/proportion of the car makes?

Answer 3

idk i just need the answer

Answer 4

Hm I feel like there is missing information.

Since the car break failures don't add to 1, I'm assuming they are conditional probabilities. So if we let \(F\) be the event that a car has a break failure (and \(A,B,C,D,E\) are the the makes of cars), then the table is giving you the probabilities: \(P(F|A), P(F|B), ... , P(F|E)\)

The question is asking you to find though, \(P(D|F\))
You could say:\[ P(D|F)=\frac{P(D\cap F)}{P(F)}\]
but we don't even know just the probability of failure, \(P(F)\). You could try using Bayes' Theorem:

\[P(D|F)=\\\frac{P(F|D)P(D)}{P(F|A)P(A)+P(F|B)P(B)+P(F|C)P(C)+P(F|D)P(D)+P(F|E)P(E)}\]
but again, we don't know just the probability of the car make for D, \(P(D)\).
You could maybe use Baye's Theorem above if there is an assumption that the proportion of car makes is the same, in that case \(P(D)=1/5\), but it's not stated.