# How are conditional probability and (in)dependent events related?

In probability theory, conditional probability is a measure of the probability of an event occurring given that another event has (by assumption, presumption, assertion or evidence) occurred.

"Probability of an event occurring given that another event has occurred." - Is this the same as saying these two events are dependent? Doesn't this mean that conditional probability is a measure for dependent events?

Two events are independent, statistically independent, or stochastically independent if the occurrence of one does not affect the probability of occurrence of the other (equivalently, does not affect the odds).

Are conditional probability and Bayes' theorem applicable to independent events? What is their role in solving problems involving independent events, if any?

I'm starting out with probability and a lot of these concepts aren't clear to me yet. What is the intuitive explanation behind the formulae?

• Bayes theorem is applicable to all combination of events. I only think that in many cases the independence will make things easy enough to do it without application of the theorem of Bayes. Nov 16 '19 at 9:35

"Is this the same as saying these two events are dependent?".

No.

If $$A$$ denotes some fixed event with $$P(A)>0$$ then it induces a new probability measure on the same collection of events. At first hand for any event $$B$$ we are interested in the map $$B\mapsto P(A\cap B)$$. However this map is not in general a probability measure because it sends the outcome set $$\Omega$$ to $$P(A)$$ and it is not excluded that $$P(A)\neq1$$. To repair this we divide by $$P(A)$$ and the function $$P_A$$ on events prescribed by $$B\mapsto P(A\cap B)/P(A)$$ is a probability measure. Instead of $$P_A(B)$$ we use a different notation: $$P(B\mid A)$$. This is how the conditional probability measure wrt event $$A$$ is "born". It answers the question:

"What is the probability that event $$B$$ occurs under the extra condition that event $$A$$ occurs?"

Independence comes in if we observe that $$P(B\mid A)$$ and $$P(B)$$ appear to be the same. That happens iff $$P(A\cap B)=P(A)P(B)$$ which means exactly that $$A$$ and $$B$$ are independent events.

What is asked above can in that case indirectly be answered with:

"Occurrence (or non-occurrence) of $$A$$ has no effect on the probability of occurrence of $$B$$ so the probability that $$B$$ occurs under this condition is still $$P(B)$$. More shortly: $$P(B\mid A)=P(B)$$.

• $P(A)>1$ isn't possible. You mean $P(A)>0$, right? Nov 16 '19 at 9:35
• @Fakemistake Yes, thank you for attending me. Repaired. Nov 16 '19 at 9:36
• I'm sorry but I don't understand the mathematical aspects of probability yet. Could you add in an intuitive explanation to the existing answer? I still can't see why conditional probability doesn't imply dependence, if we keep the formulas aside. Nov 23 '19 at 13:44
• Throw twice a fair coin. Let $H_1$ be the event that the first toss gives heads and let $H_2$ be the event that the second toss gives heads. Then we can wonder: what is the probability on $H_2$ under condition that $H_1$ occurs? We are talking about conditional independence here but will the occurrence of $H_1$ have any affect on the probability of $H_2$ to occur?... No, the coin is fair and has no memory so we get $P(H_2\mid H_1)=0.5=P(H_2)$ which means that we are also dealing with independence. So apparantly conditional probability does not imply dependence. Nov 23 '19 at 13:55
• I see, so conditional probability can be calculated for independent events too but since there is no relation between the events, it's just a multiplication of the two probabilities. Is Bayes theorem the only formula to calculate conditional probability? Nov 26 '19 at 13:33