# Conditional probability in independence and mutually exclusive events.

This thread shows that if two events are to be mutually exclusive and independent, one of them should have zero probability. I worked the following example that seems to contradict conditional probability.

We pick a real number in range [0,1]

A = event that the number is rational

B = event that the number is irrational

P(A and B) = $0$ (because they are disjoint by definition)

Also, P(A) = $0$ (from measure theory/discrete maths)

Thus P(A and B) = P(A) * P(B)

This means A and B are independent.

Note that the probability that number is irrational given that it is rational is zero. This does not satisfy conditional probability.

P(B | A) = 0 $\ne$ 1 = P(B)

P(B | A) $\ne$ P(B)

This just means that occurrence of one event (A) is affecting the probability of other (B), and not independent. Where did I go wrong?

• $\mathbb{P}(B\mid A)$ is not defined if $\mathbb{P}(A)=0$. – carmichael561 Jul 12 '16 at 23:20
• What makes this "not defined"? If I write it out, it will be 1/0, agreed. But we are dealing with probabilities here, P(B|A) must have some value between 0 and 1. – User49582934 Jul 12 '16 at 23:41
• I wish to update my comment above. If A = getting $-5$ on a fair dice throw, and B = getting any number from 1 to 6 on the same throw, then P(A)=0, P(B)=1, and P(B|A) is undefined because A cannot occur, and conditional probability on A is undefined. But in rational/irrational example, A is a valid event, and P(B | A) is a valid phrase, and I believe P(B | A) = 0 – User49582934 Jul 13 '16 at 0:26

If $\mathsf P(A)>0$, then $\mathsf P(B\mid A)\cdot\mathsf P(A) = \mathsf P(B)$ .
Even if the measure $\mathsf P(B\mid A)$ can be evaluated from the model, the consequent doesn't have to hold if the antecedent doesn't; ie: $\mathsf P(A=0)$.