Conditional probability of being in the impolite category, based on behavior According to the Etiquette Institute boys can be divided into two categories: 2/3 of them are polite and 1/3 are impolite. Polite boys will let girls enter a door first in 90% of cases while impolite buys will only do so with 20% chance, independently for each girl. I have seen that John let Julia, but not Judith, enter before him.
(a) What is the probability that John belongs to the impolite category
I've done this question but I believe I have done it incorrectly. I worked out the probability John is impolite given he lets Julia through. Then I worked out the probability he doesn't let Judith through. So I have two probabilities. I felt like I should take an average of these. But that feels incorrect. Can anyone solve this?
 A: We want to find $\mathbb P(I\mid A, B^c)$, where
\begin{align}
I &= \{\text{Josh is impolite}\}\\
A &= \{\text{Josh lets Julia enter}\}\\
B &= \{\text{Josh lets Judith enter}\}\\
\mathbb P(I) &= \frac13, \text{ the prior probability of Josh being impolite}\\
\mathbb P(A\mid I^c) &= \mathbb P(B\mid I^c) = \frac9{10}\\
\mathbb P(A\mid I) &= \mathbb P(B\mid I) = \frac1{5}.
\end{align}
So using Bayes' theorem, we compute
\begin{align}
\mathbb P(I\mid A,B^c) &= \frac{\mathbb P(A,B^c\mid I)\mathbb P(I)}{\mathbb P(A,B^c)}
\end{align}
Due to independence,
$$\mathbb P(A,B^c\mid I) = \mathbb P(A\mid I)\mathbb P(B^c\mid I) = \left(\frac9{10}\right)\left(\frac15\right)=\frac9{50},$$
whereas
\begin{align}
\mathbb P(A,B^C) &= \mathbb P(A)\mathbb P(B^c)\\ &= \mathbb P(A)^2\\
&= \left[(\mathbb P(A,I) + \mathbb P(A, I^c)\right]^2\\
& \left[ \frac{P(A\mid I)}{\mathbb P(I)} + \frac{P(A\mid I^c)}{\mathbb P(I^c)}\right]^2\\
&=\left(\frac15\cdot\frac13 + \frac9{10}\cdot\frac23\right)^2\\
&=\frac49.
\end{align}
Therefore
$$\mathbb P(I\mid A, B^c) =\frac{\frac9{50}\cdot\frac13}{\frac49} = \frac{27}{200}.$$
A: It would help to draw a tree diagram, showing firstly Polite and not Polite, then Julie and not Julie, finally Judith and not Judith.
Then you want $$p(\text{polite}|\text{Julie}\cap\text{not Judith})$$
$$=\frac{p(\text{polite}\cap\text{Julie}\cap\text{not Judith})}{p(\text{Julie}\cap\text{not Judith})}$$
$$=\frac{\frac 23\times0.9\times0.1}{\frac 23\times0.9\times0.1+\frac 13\times0.2\times0.8}=...$$
