Answer with Step-by-step explanation:
We are given that A, B and C are subsets of universal set U.
We have to prove that

Proof:
Let x
Then
and 
When
then
but 
Therefore,
but 
Hence, it is true.
Conversely , Let
but 
Then
and
When
then 
Therefor,
Hence, the statement is true.
The first one would be 120 the second one would be 68
2,3,4,1, the probability is in order from greatest to least. If they asked for least to greatest then, 1,4,3,2. So use the least to greatest one.
Answer:
Step-by-step explanation: