# Intersection of two independent events

If A and B are two independent events then P(A intersection B) = P(A).P(B)
I don't refute this but it confuses me. What is the sample space in this?
For eg: - If A is the event that we get Head while tossing a coin and B is the event that we get 2 while throwing a die, then what will we be the sample space? Is it {H, T} or {1, 2, 3, 4, 5, 6}?

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chiro
Hey Avichal.

If you have two sets A and B that are independent, then the state space for all combinations is the Cartesian product C = A X B (a belonging to A, B belonging to b, c = (a,b) belonging to C).

With regards to independence, the way you get the definition is the following:

P(A|B) = P(A and B)/P(B) [definition of conditional probability].

If A is independent from any other random variable then P(A|B) = P(A). This implies:

P(A|B) = P(A) = P(A and B)/P(B). Multiplying everything by P(B) gives

P(A and B) = P(A)*P(B) and thus proved.

Hmm ... If A and B are dependent then P(A|B) = P(A and B) / P(B)
Here what will be the sample space? For B it will be different compared to A and B. So how does it work?

mathman
The sample space is the direct product space, i.e. the elements are {h,1}, {t,1}, {h,2}, etc.

haruspex
Homework Helper
Gold Member
If you have two sets A and B that are independent, then the state space for all combinations is the Cartesian product C = A X B (a belonging to A, B belonging to b, c = (a,b) belonging to C).
Hi Chiro,
I think that's a bit confusing. Perhaps you meant this:
If you have two sets A and B that are subsets of apparently unrelated event spaces Ω1, Ω2, then in order to discuss joint probabilities etc. you must first combine the event spaces. Given their independence as spaces (not to be confused with independence of events within a space), the appropriate combination is the Cartesian product ΩC = Ω1 X Ω2. Each space has its own probability function, but they are related. A ⊂ Ω1 maps naturally to A X Ω2 ⊂ ΩC. P1(A) = PC(A X Ω2) etc. Now we can understand the joint event "A and B" as (A X Ω2) ∩ (Ω1 X B) = A X B.

chiro