# Is the answer to this question in the form of P(A|B) or P(B|A)?

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)
• find P(Truth | Training) – user29418 Mar 12 '17 at 22:56
• Great, that's what I had initially thought. However, due to Baye's Theorem, I still need to find P(Training | Truth) to find P(Truth | Training). To find P(Training | Truth) for women, could I just do 500*0.8 = 400 women tell the truth, 400* 0.7 = 280 of these truthful women have done the training? – KOB Mar 12 '17 at 22:59
• Isn't P(Training and Truth) = P(Training) x P(Truth)? So P(Training and Truth)/P(Training) = P(Truth)? – KOB Mar 12 '17 at 23:11
• no, uh, you were on the right track before. Sorry for being confusing. – user29418 Mar 12 '17 at 23:12
• only under independence. en.wikipedia.org/wiki/Independence_(probability_theory) – user29418 Mar 12 '17 at 23:17

You know that $0.7 \times 500=350$ women and $0.6 \times 400=240$ men took the training, making $590$ having taken the training in total out of $900$ in total, and $310$ not having taken the training.

You also know that $0.8 \times 900 = 720$ will tell the truth about having taken the training and so $180$ will not tell the truth

• At one extreme all those not telling the truth have not taken the training (perhaps they wish to avoid having to do it). So $590$ took the training and say they took the training; $180$ did not take the training but say they did; and $130$ did not take the training and say they did not. The the proportion of those saying they took the training who tell the truth is $\frac{590}{770} \approx 76.6\%$

• At the other extreme all those not telling the truth have taken the training (perhaps the training was not memorable). So $410$ took the training and say they took the training; $180$ took the training but say they did not; and $310$ did not take the training and say they did not. The the proportion of those saying they took the training who tell the truth is $\frac{410}{410} = 100\%$

The answer to the probability that a person who says "I took the training" is telling the truth will be in this range, but you do not have any additional information. Intuition might point you towards the lower end

Let's introduce the following notation. We have $8$ elementary events and, accordingly $8$ probabilities. Every elementary events will have three components: $$(\text{ Man / Women, Took the training / Did not take the training}, \text{Lied / Did not lie}).$$

For instance $(M,T,\overline L)$ will mean that the individual asked is a man, he took the course and he did not tell a lie.

$$\begin{matrix} \text{ e. event }&\text{probability}\\ \\ (M,T,L)&p_1\\ (M,T,\overline L)&p_2\\ (M,\overline T,L)&p_3\\ (M, \overline T,\overline L)&p_4\\ (W,T,L)&q_1\\ (W,T,\overline L)&q_2\\ (W,\overline T,L)&q_3\\ (W, \overline T,\overline L)&q_4 \end{matrix}$$

We have the following information

\begin{align} p_1+p_2+p_3+p_4=\frac49,& \ \text{ (from the number of men)}\\ q_1+q_2+q_3+q_4=\frac59,& \ \text{ (from the number of women)}\\ p_1+p_3+q_1+q_3=\frac8{10},&\ \text{ (from the number of liars)}\\ p_1+p_2=\frac49\frac6{10},&\ \text{ (from the number of men having taken the training)}\\ q_1+q_2=\frac59\frac7{10},&\ \text{ (from the number of women having taken the training)}.\\ \end{align}

From these equations the individual probabilities cannot be determined.

So we cannot compute the conditional probabilities asked by the OP:

$$P(\text{"Told the truth"}\ \mid \ \text{"Took the training"})=\frac{p_2+q_2}{p_1+p_2+q_1+q_2},$$

$$P(\text{"Took the training"}\ \mid \ \text{"Told the truth"})=\frac{p_2+q_2}{p_2+p_4+q_2+q_4}.$$

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

$$P(T|B) = \frac{\frac{590}{900}\times 0.8}{\frac{590}{900}\times 0.8 + (1-\frac{590}{900}) \times 0.2}$$