Mathematics, statistics 150

Question

 If A and B are two mutually exclusive events with P( A) = 0.35 and

P(B)=0.55 , find the following probabilities: 

 (a) P( A and B)=

(b) P( A or B)=

(c) P(not A) =

(d) P(not B)=

(e) P(not (A or B))=

(f) P(A and (not B))=

Answer 1

(a) P( A and B)=P(A)*P(B) = (0.35)(0.55) = 0.1925

(b) P( A or B)=P(A)+P(B) = (0.35)+(0.55) = 0.90

(c) P(not A) = 1-P(A) = 1-0.35 = 0.65

(d) P(not B)=1-P(B) = 1-0.55 = 0.45

(e) P(not (A or B))=1-P((A or B)) = 1-0.90 = 0.10

(f) P(A and (not B))=P(A)*P(not B) = (0.35)(0.45) = 0.1575

Answer 2

Two events are said to be mutually exclusive if they can’t both happen at the same time.

So, P(A and B) = 0 { As P(A and B) means both A and B occurring at the same time }

And P(A or B) which means P(A) + P(B) - P(A and B) As P(A and B) = 0. which implies

P(A or B)= P(A) + P(B)

Note down : For two Mutually Exclusive Events :

1) P(A and B) = 0 and

2) P(A or B)= P(A) + P(B)

(a) P( A and B)= 0

(b) P( A or B)= P(A) + P(B) = 0.35+0.55 = 0.9

(c) P(not A) = 1 - P(A) => 1-0.35 = 0.65

(d) P(not B)= 1 - P(B) => 1-0.55 = 0.45

(e) P(not (A or B))= 1 - P(A or B) = 1- 0.9 = 0.1

(f) P(A and (not B))= P(A) - P(A and B) = 0.35 - 0 = 0.35

If you have any further doubts, you can message me, i am available anytime to clarify.