Both scenarios you presented are of mutually exclusive sets. So for the first scenario the Probability of the intersection would be 0
The second scenario 19/105 + 52/105 since again they will not have an intersection you will not need to subtract the elements they have in common
Close, you forgot to add one (twice).
The first scenario is Mut Ex.
To get all multiples of 2 (inclusive) ---> 105-1/2 = 52
To get all numbers ending with 3 (inclusive) ---> 103-3/10+1
To get all numbers ending with 7 (inclusive) ---> 97-7/10+1
Mut Ex. Formula: Prob(even)+Prob(3)+Prob(7)/total possibilities
Simplify to 73/105 ~70%
Now how could we rewrite this problem so that it wouldn't
be Mut Ex?