Three friends A, B and C got their photograph clicked. Find the probability that B is standing at the central position, given that A is standing at the left corner.
Now, This is obviously a case of conditional probability so if the event that B is in the centre is $E_1$ and the event that A is on the left is $E_2$ then
$ P(E_1|E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)} $which is $\frac{1/2}{1/3}=\frac{3}{2}$
which is obviously wrong
where am I messing up?
On
Let the event that $B$ is in the middle be $B_{mid}$ and the event that $A$ is at the left corner be $A_{left}$. So what you are trying to find is $P(B_{mid}| A_{left})$. By conditional probability we have, $$P(B_{mid}| A_{left})=\frac{P(B_{mid}\cap A_{left})}{P(A_{left})} $$
Consider the event $B_{mid}\cap A_{left}$, the only possible arrangement of the friends here is $ABC$. So the probability of the arrangement of $ABC$ happening out of the total number of arrangements is $\frac{1}{3!}$. So now we know that $P(B_{mid}\cap A_{left}) = \frac{1}{3!}$.
Consider the event $A_{left}$, what arrangements are there such that $A$ is at the left corner? They are $A**$ where each $*$ can be $B$ or $C$, basically $ABC$ and $ACB$. So the number of arrangements where $A$ is at left corner is $2!$. Similarly we can conclude that $P(A_{left}) = \frac{2!}{3!}$.
Putting all information together we conclude that $$P(B_{mid}| A_{left})=\frac{\frac{1}{3!}}{\frac{2!}{3!}} = \frac{1}{2}$$
Since $E_1\cap E_2$ is a subset of $E_2$, you should have $P(E_1\cap E_2) < P(E_2)$.
In other words, whenever $E_1\cap E_2$ is satisfied, $E_2$ is also satisfied.
Just by writing out the options: