Basic formula of Probability I was wondering how probability of an event is equal to favourable cases/total cases
like on tossing a coin probaility of getting a head is 1/2 so why is it alwars favourable cases/total cases Does this formula or law has any proof.
 A: The (# favorable outcomes)/(# total outcomes) rule does not always hold. It is only true when there are finitely many outcomes, and these outcomes are all equally likely. Given these two assumptions, here is how you prove that the rule works:
Theorem: In a probability space $\Omega$ with finitely many outcomes where all outcomes are equally likely, $P(E)=|E|/|\Omega|$ for any event, $E$.
Proof: Let $n=|\Omega|$, and let $\omega_1,\omega_2,\dots,\omega_n$ be the complete list of outcomes.
Let $p$ be the probability of any particular outcome. Using the axiom $P(\Omega)=1$, we have
$$
1=P(\Omega)=P(\{\omega_1,\dots,\omega_{n}\})\stackrel{*}=P(\{\omega_1\})+\dots+P(\{\omega_n\})=p+\dots+p=np
$$
We can then solve for $p$, getting $p=1/n$. Finally, for any event $E$ with $k$ outcomes $e_1,\dots,e_k$, we have
$$
P(E)=P(\{e_1,\dots,e_k\})\stackrel{*}=P(\{e_1\})+\dots+P(\{e_k\})=\frac1n+\dots+\frac1n=\frac kn=\frac{|E|}{|\Omega|}
$$
The claim has been proven. To be crystal clear, in the $\stackrel{*}=$ equations, we use the probability axiom $P(A\cup B)=P(A)+P(B)$ whenever $A$ and $B$ are disjoint events.  This further implies that $P(A_1 \cup \dots \cup A_n)=P(A_1)+\dots+P(A_n)$ whenever the events $A_1,\dots,A_n$ are pairwise disjoint, which can be proven by induction on $n$.

The assumption of all outcomes being equally likely is essential. For example, say you had a loaded die where the six face is more likely to come up than the other faces (because the opposite face, one, has the fewest pips, so is heaviest). Specifically, suppose the probability of rolling six is $25\%$, and the probability of rolling any other face is $15\%$. To compute the probability of the event of rolling an even number, the rule (# good outcomes)/(# total outcomes) would give $3/6$, which is incorrect. The correct answer is $$P(\text{even})=P(\{2,4,6\})=P(\{2\})+P(\{4\})+P(\{6\})=0.15+0.15+0.25 = 0.55$$
