TITLE: Is the answer to this question in the form of P(A|B) or P(B|A)?
QUESTION [0 upvotes]: I am trying to solve the following question using Bayes' Theorem:
In a company of 500 women and 400 men, the records of which employees had already completed a mandatory training class were lost. All that was now known is that 70% of women and 60% of men had already taken the training. If only 80% of employees told the truth when asked if they had taken the training, what is the probability that a person who says "I took the training" is telling the truth?
Firstly, can I divide the question into two separate calculations, men and women, and then combine them using the Law of Total Probability to find the complete answer?
Secondly, is the question asking to find the probability that

a person is telling the truth, given that they have done the training? - P(Truth | Training)
a person has done the training, given that they are telling the truth? - P(Training | Truth)

REPLY [0 votes]: The person who said I did the training has either did it and is telling the truth or did not trained and lying.
So: 
let B: the person says that trained
let T: the person is telling the truth
let V: the person trained
$P(B) = P(V \cap T) + P(notV \cap notT)$
so B: trained and tells the truth plus not trained and lie
$P(T|B) = \frac{P(T \cap B)}{P(B)}$
$P(T|B) = \frac{P(T \cap V)}{P(V \cap T) + P(notV \cap notT)}$
Yo can supose that telling the truth is independent of training or sex because there is no information that says otherwise
$P(V) = \frac{0.7 \times 500 +0.6 \times 400}{900} = 590/900$
$P(T) = 0.8 $
so 
$P(T|B) = \frac{\frac{590}{900}\times 0.8}{\frac{590}{900}\times 0.8 + (1-\frac{590}{900}) \times 0.2}$
{"set_name": "stack_exchange", "score": 0, "question_id": 2183971}