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I have this question in exercise which I solved it but not sure about the answer.

Question

There are two department in a car manufacturing company, each department is making cars. Department 1 has 10 racing cars and 30 luxury cars, while department 2 has 20 of each. A marketing manager picks a department at random, then picks a car at random. The car turns out to be a luxury car. How probable is it that the manager picked it out of department 1?

Answer

Probability of picking a luxury car from department 1 $= 1/2*(30/40)=0.375$

Probability of picking a luxury car from department 2 $= 1/2*(20/40)=0.25$

More probable than picking up from department 2.

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  • $\begingroup$ Let $A$ be the event picked from 1, $B$ the event picked luxury car. We want $\Pr(A|B)$, which by definition is $\frac{\Pr(A\cap B)}{\Pr(B)}$. Can you find the two required probabilities? $\endgroup$ – André Nicolas Jan 25 '15 at 5:31
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Use Bayes' rule:

P(Dep1|Rac) = P(Rac|Dep1)P(Dep1)/(P(Dep1)P(Rac|Dep1) + P(Dep2)P(Rac|Dep2))

where

P(Dep1|Rac): probability the state of the world is Dep1 given a racing car was selected--unknown

P(Rac|Dep1): probability a racing car is selected given the department is Dep1--0.25

P(Dep1): prior probability Dep1 is chosen--0.5

P(Rac|Dep1): probability of choosing a racing car given it is from Dep1--0.25

P(Dep2): prior probability Dep2 is chosen--0.5

P(Rac|Dep2): probability of choosing a racing car given it is from Dept2--0.5

Substituting and doing the arithmetic gives

P(Dep1|Rac) = 1/3.

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