1

In my maths schoolbook there is this property for independent events:

p(A|B) = p(A)

p(B|A) = p(B)

But I can't imagine how this is possible. How would you draw a Venn diagram that follows this property?

Segmentation fault
  • 1,873

2 Answers2

3

Here is a diagram that illustrates independence, where probability means "area". Here $P(A)$ is the proportion of the square that is colored yellow, while $P(A\mid B)$ is the proportion of the green stripe that is colored yellow. These two proportions are the same (approximately 1/3).

enter image description here

1

The events $A$ and $B$ are independent if $$ P(A\cap B)=P(A)P(B). $$ The conditional probability of the event $A$ given the event $B$ is defined by $$ P(A\mid B)=\frac{P(A\cap B)}{P(B)} $$ provided that $P(B)>0$. So we have that $$ P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A)P(B)}{P(B)}=P(A) $$ if $A$ and $B$ are independent. Analogously, $P(B\mid A)=P(B)$. So we see that this property holds if and only if the events $A$ and $B$ are independent. This property can be taken as a definition of the independence.

Cm7F7Bb
  • 17,879
  • 5
  • 43
  • 69
  • But how is this possible? How is it possible that P(something)/P(B) = P(something)? In my view, this can only happen if P(B) = 1, right? – Segmentation fault Sep 30 '16 at 13:41
  • Wait, actually no, because multiplying P(B) and dividing by it cancel each other. – Segmentation fault Sep 30 '16 at 13:43
  • It is not like that. The property states that if two events $A$ and $B$ are independent, then $P(A\cap B)/P(B)=P(A)$. But this is true because $P(A\cap B)=P(A)P(B)$ if $A$ and $B$ are independent. – Cm7F7Bb Sep 30 '16 at 13:43