Question

If C1 and C2 are subsets of the same sample space C, show that P(C1 inersection C2)<=P(C1)<=P(C1 U C2)<=P(C1) + P(C2)

Answer 1

Clearly, the intersection of any two of these sets is the null set, which has probability 0.

For example, with the first two sets, C1 intersection C2 and C1 intersection ~C2, their intersection is C1 intersection(C2 intersection ~ C2) = C1intersection the null set is the null set.

a, b, c, and d are all greater than or equal to 0.

When the intersection of two sets is the null set, the probability of their union is the sum of their respective probabilities.

Henceforth, I will use n to stand for intersection and U to stand for intersection within the set notation.

Then, P(C1 n C2) = a (given above).

C1 = (C1 n C2) U (C1 n ~C2) (If you are in C1, you have to either be in C2 or not be in C2). In set notation, you can write C1 = C1 nΩ = C1 n (C2 U ~ C2) = C1nC2 U C1n ~C2

Then, as (C1 n C2)and (C1 n ~C2) have the null set as their intersection,

P(C1) = a + b.

C1 U C2 = C1 n (~C1 n C2) = (C1 n C2) U (C1 n ~C2) U (~C1 n C2)

The intersection of any two of these sets is the null set. Therefore,

P(C1UC2) = a + b + c.

C2 = (C1 n C2)U (~C1 n C2)

Then, P(C2) = a + c.

P(C1) + P(C2) = a+b + a+c = 2a+b+c

As a, b, and c are non-negative,

a <= a+b <= a+b+c <= 2a+b+c (the first relationship, a <= a+b, is true because b >=0; the second relationship, a+b <=a+b+c, is true because c >= 0; the third relationship, a+b+c <= 2a+b+c, is true because a >= 0).

Then, we may substitute.

P(C1 n C2) <= P(C1) <= P(C1 U C2) <= P(C1) + P(C2)

I hope this meets your needs.