0
$\begingroup$

Regarding this question and its highest-voted answer: Conditional probability intuition.

So I get the idea of using Venn-Diagrams and limiting our "universe" to a new subset, but what happens if we try looking at it with a tree diagram?
Using this equation: $$P(A\vert B)P(B)=P(B\vert A)P(A)$$ we can set the probabilities to whatever we'd like, for example let's say $P(A)=\dfrac12$, $P(B)=\dfrac13$, $P(A|B)=\dfrac14$, $P(B|A)=\dfrac15$.
The equation does not hold, the events are dependent on each other.
If sequence of occurrence does matter, how can you continue using a Venn-Diagram to prove this equality?

$\endgroup$
2
$\begingroup$

You can't set probabilities to whatever you like. $P(A|B)$ is defined as being $P(A,B)/P(B)$ and $P(B|A)$ is defined as being $P(A,B)/P(A)$. If you have $P(A)=1/2$ and $P(B)=1/3$ and $P(A|B)=1/4$ then $P(A,B)=1/12$ and $P(B|A)=1/6$. You can't set $P(B|A)=1/5$ any more than you can set $2+2=5$.

$\endgroup$
1
$\begingroup$

There is a problem with the probabilities that you have assigned: $$ \mathbb{P}(A|B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} \implies \mathbb{P}(A\cap B) = \frac{1}{12} $$ On the other hand, $$ \mathbb{P}(B|A) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(A)} \implies \mathbb{P}(A\cap B) = \frac{1}{10} $$ You can set the probabilities whatever you would like, but once you have specified any 3 of the following: $\mathbb{P}(A)$, $\mathbb{P}(B)$, $\mathbb{P}(A|B)$, $\mathbb{P}(B|A)$ or $\mathbb{P}(A \cap B)$ the remaining two will be determined.

The proof of the equality is a simple corollary from the definition of conditional probability: $\mathbb{P}(A|B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}$ and hence it follows that the occurrence of events should not matter.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.