Relation of Bayes Theorem, Conditional Probability and Likelihood Bayes Theorem is derived using conditional probability:
$$P(A|B)\cdot P(B) = P(A) \cdot P(B|A)$$
Here terms are either probability or conditional probability.
But later, in
$$P(A|B) = P(A)\cdot \frac{P(B|A)}{P(B)}$$
$P(B|A)$ is called likelihood.
Since likelihood and probability are not the same. Please help me understand: How did we end up with likelihood here and when do we use conditional probability and when do we use likelihood?
 A: As you note, $P(B|A)$ plays two roles. The role depends on whether $A$ is fixed or $B$ is fixed.
When $A$ is fixed, $P(B|A)$ is the probability of $B$ given the value of $A$. Since probabilities add up to one, we have $\sum_B P(B|A) = 1$ for any value of $A$ for which the conditional probability is defined.
When $B$ is fixed, $P(B|A)$ is a function of $A$. This function is called the likelihood. Since likelihood is not probability, $\sum_A P(B|A)$ does not have to equal 1. It doesn't even have to be finite. Nevertheless, $\sum_A P(B|A)\,P(A) = P(B)$, which depends on $B$.
$B$ is fixed when Bayes theorem is used to compute the conditional probability $P(A|B)$:
\begin{equation}
P(A|B) = \frac{P(B|A)\,P(A)}{P(B)} .
\end{equation}
Therefore, $P(B|A)$ plays the likelihood role in this expression.
If instead we used Bayes theorem to compute the probability $P(B|A)$, the roles would be reversed and $P(A|B)$ would play the likelihood role:
\begin{equation}
P(B|A) = \frac{P(A|B)\,P(B)}{P(A)} .
\end{equation}
