In Bayes’ formula:

P(x|a)=P(a|x)P(x)P(a)

can the posterior probability P(x|a) exceed 1?

I think it is possible if for example, assuming that 0<P(a)<1, and P(a)<P(x)<1, and P(a)/P(x)<P(a|x)<1. But I'm not sure about this, because what would it mean for a probability to be greater than one?

**Answer**

The assumed conditions do not hold- it can never be true that P(a)/P(x)<P(a|x) by the definition of conditional probability:

P(a|x)=P(a∩x)/P(x)≤P(a)/P(x)

**Attribution***Source : Link , Question Author : Thomas Moore , Answer Author : khol*