Probability and Independence: Proving Subset Elements with Uniform Distribution

Proof of Independence for Subsets with Uniform Distribution

Let Ω = {1, 2, 3, 4, 5, 6} with uniform probability. If A ⊂ Ω and B ⊂ Ω are independent and A has four elements, then B must have 0, 3, or 6 elements.

To prove this, we start by noting that the probability of the intersection of A and B is given by:

P(A ∩ B) = P(A) * P(B)

Since A and B are independent, we have:

P(A ∩ B) = P(A) * P(B) = (4/6) * P(B) = (2/3) * P(B)

Now, since P(A ∩ B) is a probability, we have:

0 ≤ P(A ∩ B) ≤ 1

Therefore, we have:

0 ≤ (2/3) * P(B) ≤ 1

This implies that:

0 ≤ P(B) ≤ 3/2

Since the number of elements in B must be a non-negative integer, we conclude that:

0 ≤ |B| ≤ 3

However, since B is a subset of Ω, we also know that:

|B| ≤ 6

Combining the above two inequalities yields:

0 ≤ |B| ≤ 3 or |B| = 6

Therefore, B must have 0, 3, or 6 elements, as required.

QED.