Question

Let A , B , and C be disjoint subsets of the sample space. For each one of the following statements, determine whether it is true or false. Note: "False" means "not guaranteed to be true."

a) P(A)+P(Ac)+P(B)=P(A∪Ac∪B)

b) P(A)+P(B)≤1

c) P(Ac)+P(B)≤1

d) P(A∪B∪C)≥P(A∪B)

e) P((A∩B)∪(C∩Ac))≤P(A∪B∪C)P((A∩B)∪(C∩Ac))≤P(A∪B∪C)

f) P(A∪B∪C)=P(A∩Cc)+P(C)+P(B∩Ac∩Cc)

)

Please explain how you got the answer.

Answer 1

a.)

P(A)+P(Ac)+P(B)=P(A∪Ac∪B)

Taking RHS,

P(A∪Ac∪B) = P(A) + P(Ac) + P(B) +P(A Ac) - P(A Ac) - P(A B) - P(B Ac) + P(A B Ac)

Now, A Ac = NULL, P(A Ac) = 0

Since A and B are disjoint, So, A B = NULL, P(A B) = 0

Now For Ac B

From above diagram,

Ac B = B, So, P(Ac B) = P(B)

So, Ac B A = NULL, P(Ac B A) = 0

This results in,

P(A∪Ac∪B) = P(A) + P(Ac) + P(B) - 0 - 0 - P(B) - 0

P(A∪Ac∪B) = P(A) + P(Ac) which is not equal to Left hand side(LHS) Hnece above statement is False

b.) P(A) + P(B) 1

The Probability of whole Sample Space is equal to 1.

-> If we don't consider C and complement of A and B, the sum of probabilities is always less than 1.

The below image proves the above statement

-> In case if P(C) = 0, and whole sample constitutes of only A and B. Then,

P(A) + P(B) = 1

So, above 2 statements prove that P(A) + P(B) 1. Hence it is True.

c.) P(Ac) + P(B) 1

The above statement is not always True. Consider below 2 examples,

-> Case when P(A) > P(B)

Let's say P(A) = 0.2, P(B) = 0.1, P(C) = 0.3

So, P(Ac) = 1 - 0.2 = 0.8

P(Ac) + P(B) = 0.9 which is less than 1. Hence above statment hold true.

-> Case when P(A) < P(B)

Let's say P(A) = 0.1, P(B) = 0.3, P(C) = 0.2

So, P(Ac) = 1 - 0.1 = 0.9

P(Ac) + P(B) = 1.2 which is more than 1. Hence above statment hold False.

Hence statement P(Ac) + P(B) 1 is False.

d.)

P(AUBUC) P(AUB)

Since A, B and C are disjoint so by property,

P(AUBUC) = P(A) + P(B) + P(C) --------(1)

P(AUB) = P(A) + P(B) --------(2)

From above 2 equations,

P(AUBUC) is always greater than P(AUB). In Case if P(C) = 0, then P(AUBUC) is equals to P(AUB)

Hence, statement P(AUBUC) P(AUB) is always TRUE