We have four coins in a box, let $p_i$ describe the probability of getting head from coin $i$: $p_1=0, p_2=0.25, p_3=0.5, p_4=0.75$.
We toss take one coin and toss it until we get head, what is the probability of tossing it exactly 3 times?
I can calculate it fairly easily with a tree to get: $0.25(0.25\cdot0.75^2+0.5^3+0.25^2\cdot0.75)=0.078125$
But why we don't have to divide that result by $|\Omega|=2^3$ (heads or tails 3 times) or by the probability of getting tails in the first two tries?
I.e: $A= \text{heads in the third try} \\ B=\text{not head in the first and second tries}$
$P(A|B)=\frac {P(A\cap B)}{P(B)}=\frac{0.078125}{0.25(0.75^2+0.5^2+0.25^2)}=0.357143$
I'm basically asking why is the answer $P(A \cap B)$ and not $P(A|B)$ or $P(A \cap B)/|\Omega|$ ?