Possible bayes theorem question regarding a shared car The question: Suppose 2 friends share the use of a car evenly, we will call the two friends Roger and James, respectively. We know that Roger will only use the car to drive to the grocery store, on the other hand when James uses the car, 1/3rd of the time he will drive to the grocery store.
If we were to see Roger and James car parked outside the grocery store with nobody inside, what are the odds that Roger is inside the grocery store ?
Solution:
My logic is to draw a tree diagram and focus only on the possibilities that result in Roger at the grocery store, out of the total that includes both Roger and James, which so happens to be 3/4, can anyone confirm if this is correct ?
 A: Alternative approach:
The problem can be informally (visually) attacked intuitively without (much) math, by initially deciding that there are $6$ equally likely possibilities:

*

*$3$ of which Roger drives - all $3$ times, he goes to the store.


*$3$ of which James drives - exactly one of the $3$ times, he goes to the store.
So, you (presumably) initially have $6$ equally likely possibilities.  Then $2$ of these $6$ possibilities are eliminated because you are given that the car is in the store parking lot.
So, you are left with only $4$ equally likely  possibilities, only $1$ of which involves James driving to the store.

The motivation here is that it is necessary to divide the event of James having the car into $3$ sub-events, only one of which has him driving to the store.  Further, you have the presumption that absent any other information, Roger and James are equally likely to be the one using the car.
Therefore, since $3$ sub-events were assigned to James, you must balance the sub-events by also assigning $3$ sub-events to Roger.
A: That is a valid solution.  To confirm you may use Bayes' Rule (and the Law of Total Probability).
Using the events of $R,J,$ and $G$ to be "Roger took the car", "James took the car", and "the car was taken to the grocers", we have:  $$\def\P{\operatorname{\sf P}}\P(R\mid G) =\dfrac{\P(G\mid R) \P(R)}{\P(G\mid J) \P(J)+\P(G\mid R) \P(R)}$$
Now, we are given: $\def\P{\operatorname{\sf P}}\P(J)=\P(R), \P(G\mid R)=1, \P(G\mid J)=1/3$ so....
