# Thread: Elementary question regarding conditional probability and independence

1. ## Elementary question regarding conditional probability and independence

A ball is drawn at random from a box containing 10 balls numbered sequentially from 1 to 10. Let X be the number of the ball selected, let R be the event that X is an even number, let S be the event that $\displaystyle X\geq 6$, and let T be the event that $\displaystyle X\leq 4$. Which of the pairs (R,S), (R,T), and (S,T) are independent?

I can solve this mathematically as such:
$\displaystyle \newline P(R)=0.5\newline P(S)=0.5\newline P(T)=0.4$

$\displaystyle P(R\cap S)=P\{6,8,10\}=0.3\neq P(R)*P(S)=0.25$

$\displaystyle P(R\cap T)=P\{2,4\}=0.2= P(R)*P(T)=0.20$

$\displaystyle P(S\cap T)=P\{\emptyset\}=0\neq P(S)*P(T)=0.20$

So $\displaystyle P(R\cap T)$ is independent.

My question is, intuitively, how is $\displaystyle P(R\cap T)$ independent, but $\displaystyle P(R \cap S)$ not? Maybe I'm not understanding what to events being independent entails.

2. ## Re: Elementary question regarding conditional probability and independence

Hey downthesun01.

Independence intuitively means that if you have information for one event that it doesn't effect the probability for another event.

So consider the two events: an even number and a number <= 4. If you know that something is even (2,4) then given that its also <= 4 you don't have any extra advantage of knowing what number it is. If its even then you have an equal chance of picking between 2 and 4 and if its odd again you have an equal chance of picking between 1 and 3.

No matter what the scenario, knowing one event doesn't help in predicting another and this is exactly what independence means intuitively in a probabilistic sense.

If one event did greatly effect the probability then you would get an P(A and B) != P(A)P(B).

The proof for independence is really simple. If B and A are independent then P(A|B) = P(A) and P(B|A) = P(B). Since P(A|B) = P(A and B)/P(B) = P(A) we multiply both sides by P(B) and get P(A|B) = P(A)*P(B) for independent events.