I am trying to solve problem 5 (https://www.probabilitycourse.com/chapter1/1_4_5_solved3.php). Part a) and b) I got right, part c) an extra factor appears in there for me. Here's my work below. Any ideas what I'm getting wrong/why I might be getting it wrong?
In my town, it's rainy one third of the days. Given that it is rainy, there will be heavy traffic with probability $\frac{1}{2}$, and given that it is not rainy, there will be heavy traffic with probability $\frac{1}{4}$. If it's rainy and there is heavy traffic, I arrive late for work with probability $\frac{1}{2}$. On the other hand, the probability of being late is reduced to $\frac{1}{8}$ if it is not rainy and there is no heavy traffic. In other situations (rainy and no traffic, not rainy and traffic) the probability of being late is 0.25. You pick a random day.
Given that I arrived late at work, what is the probability that it rained that day?
Givens:
$P(R) = \frac{1}{3}$
$P(T|R) = \frac{1}{2}$
$P(T|R^c) = \frac{1}{4}$
$P(L|R,T) = \frac{1}{2}$
$P(L|R^c,T^c) = \frac{1}{8}$
$P(L|R,T^c) = P(L|R^c,T) = \frac{1}{4}$
$P(L) = \frac{11}{48}$ - from question b)
$P(R,T) = P(R,T^c) = \frac{1}{6}$
To answer the question I need to solve $P(R|L)$. I use Bayes
$P(R|L) = \frac{P(L|R)P(R)}{P(L)}$
I have $P(R)$ and $P(L)$. So to solve $P(L|R)$, I use the Law of Total Probability and add conditioning on T.
$P(L|R)$ = $P(L|R,T)P(R,T) + P(L|R,T^c)P(R,T^c)$ = $\frac{1}{2}.\frac{1}{6} + \frac{1}{4}.\frac{1}{6} = \frac{1}{48}$
Substituting the values, I get $P(R|L) = \frac{\frac{1}{8}.\frac{1}{3}}{\frac{11}{8}}$.
But this isn't correct, somehow, I introduce an extra $\frac{1}{3}$, i.e. the $P(R)$. Without it, the result would be correct.
The relation you are using to calculate $P(L|R)$ is wrong. Here's what it should be $^{[1]}$
$$\begin{align*} P(L|R) &= P(L|R,T) \cdot P(T|R) + P(L|R,T^c) \cdot P(T^c|R) \\[0.3cm] &= \frac{3}{8} \end{align*}$$
So,
$$P(R|L) = \frac{\frac{3}{8} \cdot \frac{1}{3}}{\frac{11}{48}} = \frac{6}{11}$$
$^{[1]}$ On the left side, "raining" is a given. So, it must be so on the right side too. $P(R, T)$ is the probability of the day being rainy and traffic being heavy. We need the probability of the traffic being heavy given it's rainy $P(T | R)$.
Similarly, the other term in the sum would be $P(T^c | R)$.