# Conditional Probability between two events

There are two events A and B. Whenever event B occurs the event A will also occur. But it is not necessary that whenever event A occurs, the event B will also occur.

Now I want to calculate the probability of B given that event A has already occurred, i.e. P(B|A).

Total number of events for B = n(B) Total number of events for A = n(A)

I guess the probability of P(B|A) should be equal to n(B)/n(A). But I am not sure as I am not good at probability.

• The definition of probability of $B$ being $Pr(B)=n(B)/n(S)$ will only take you so far and only works when talking about a finite equiprobable sample space. You should get out of the habit of thinking in such limited ways. Hint to continue: Look at Bayes' theorem. – JMoravitz Apr 11 '17 at 17:35
• @JMoravitz I looked at Bayes rule, according to it P(B|A) = (P(A|B).P(A))/(P(A|B).P(B) + P(A|not B)).P(not B)). In my scenario, solving the denominator, P(A|B) will be 1, P(A| not B) will become 0 as A cannot occur without B. This gives us P(B) in the denominator and P(A) in the numerator. So my final equation will look like P(B|A) = P(A)/P(B). Not sure, whether this right. Because if P(A)>P(B) then the probability will become > 1. – John Rambo Apr 11 '17 at 17:59
• Bayes' rule says that $P(B\mid A)=P(A\mid B)P(B)/P(A)$. The version you looked at would be if you didn't know what $P(A)$ was equal to, but you are implying that you do know., Next, note that the phrase "whenever B occurs A will occur" implies $P(A\mid B)=1$, so the above simplifies to $P(B)/P(A)$. ONLY if you are working in a finite equiprobable sample space can you then simplify that further to $n(B)/n(A)$, since it is generally not true that $P(B)/P(A)=n(B)/n(A)$, I would leave it as $P(B)/P(A)$ and not simplify further. – JMoravitz Apr 11 '17 at 18:03
• Also in what you wrote you seem to have mixed around your B's and A's in a few small places., the numerator should have been $P(A\mid B)P(B)$, not $P(A\mid B)P(A)$ – JMoravitz Apr 11 '17 at 18:06
• Thanks!! I realised my mistake. – John Rambo Apr 11 '17 at 18:07

Hint: $P(B|A)=\dfrac {P(A\cap B)}{P(A)}=\dfrac {P(B)}{P(A)}\cdot P(A|B)$
• Yes, $P(A|B)=1$. Look at Jmoravitz comment about $\frac{n(B)}{n(A)}$, if you have a finite equiprobable sample space then you are totally right – Giulio Apr 11 '17 at 17:47