Step-by-step explanation:
We have to prove both implications of the affirmation.
1) Let's assume that A \ B and C \ D are disjoint, we have to prove that A ∩ C ⊆ B ∪ D.
We'll prove it by reducing to absurd.
Let's suppose that A ∩ C ⊄ B ∪ D. That means that there is an element x that belongs to A ∩ C but not to B ∪ D.
As x belongs to A ∩ C, x ∈ A and x ∈ C.
As x doesn't belong to B ∪ D, x ∉ B and x ∉ D.
With this, we can say that x ∈ A \ B and x ∈ C \ D.
Therefore, x ∈ (A \ B) ∩ (C \ D), absurd!
It's absurd because we were assuming that A \ B and C \ D were disjoint, therefore their intersection must be empty.
The absurd came from assuming that A ∩ C ⊄ B ∪ D.
That proves that A ∩ C ⊆ B ∪ D.
2) Let's assume that A ∩ C ⊆ B ∪ D, we have to prove that A \ B and C \ D are disjoint (i.e. A \ B ∩ C \ D is empty)
We'll prove it again by reducing to absurd.
Let's suppose that A \ B ∩ C \ D is not empty. That means there is an element x that belongs to A \ B ∩ C \ D. Therefore, x ∈ A \ B and x ∈ C \ D.
As x ∈ A \ B, x belongs to A but x doesn't belong to B.
As x ∈ C \ D, x belongs to C but x doesn't belong to D.
With this, we can say that x ∈ A ∩ C and x ∉ B ∪ D.
So, there is an element that belongs to A ∩ C but not to B∪D, absurd!
It's absurd because we were assuming that A ∩ C ⊆ B ∪ D, therefore every element of A ∩ C must belong to B ∪ D.
The absurd came from assuming that A \ B ∩ C \ D is not empty.
That proves that A \ B ∩ C \ D is empty, i.e. A \ B and C \ D are disjoint.
. In this case our desired out come is all multiples of 3: 3,6,9 and 12. The total number of outcomes can be found as 100 by adding up all the frequencies. Thus our experimental probability is 
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