A problem in probability goes like:
If you have, say three similar cups, one of which holds a marble (the cups are opaque and inverted so you can't see the inside of the cups), the probability of finding the marble in any one of them is 1/3. Name the cups A, B, and C for convenience. Now if you were asked to choose the cup that holds the marble and you seemingly choose B, but then you are told that the marble is not in A, then if you're asked to keep your choice or switch to C. It is known that switching, certainly is a better option, because the odds of finding the marble in C are 2/3 while in A, they are 0, and 1/3 in B. Which can be proved using the fact that, $\mathsf P(B)+(P(A)+P(C))=1.$
Can this be proved using conditional probability?
Let $A_m, B_m$ and $C_m$ be the events that the marble is in cups $A, B$ and $C$ respectively; also let $A_r$ be the event that cup $A$ is revealed, i.e. shown not to contain the marble.
Suppose you choose cup $B$, and are then shown that the marble is not in cup $A$. Then the probability of winning by keeping cup $B$ is
$$\begin{align} P(B_m | A_r) &= \frac{P(B_m \cap A_r)}{P(A_r)} \\[2mm] &= \frac{P(A_r | B_m) \cdot P(B_m)}{P(A_r | A_m) \cdot P(A_m) + P(A_r | B_m) \cdot P(B_m) + P(A_r | C_m) \cdot P(C_m)} \\[2mm] &= \frac{\left( \frac{1}{2} \right) \left( \frac{1}{3} \right)}{\left( 0 \right) \left( \frac{1}{3} \right) + \left( \frac{1}{2} \right) \left( \frac{1}{3} \right) + \left( 1 \right) \left( \frac{1}{3} \right)} \\[2mm] &= \frac{1}{3} \end{align}$$
Similarly, the probability of winning by switching to cup $C$ is
$$\begin{align} P(C_m | A_r) &= \frac{P(C_m \cap A_r)}{P(A_r)} \\[2mm] &= \frac{P(A_r | C_m) \cdot P(C_m)}{P(A_r | A_m) \cdot P(A_m) + P(A_r | B_m) \cdot P(B_m) + P(A_r | C_m) \cdot P(C_m)} \\[2mm] &= \frac{\left( 1 \right) \left( \frac{1}{3} \right)}{\left( 0 \right) \left( \frac{1}{3} \right) + \left( \frac{1}{2} \right) \left( \frac{1}{3} \right) + \left( 1 \right) \left( \frac{1}{3} \right)} \\[2mm] &= \frac{2}{3} \end{align}$$