Question

Discuss and criticze the following
P(A)=2/3, P(B)=1/4 and P(C)= 1/6 for the probability of three mutually exclusive
Eventy A, B and C​

Answer 1

Answer:

Step-by-step explanation:

Since A, B and C are mutually exclusive events,

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

                  = 2/3 + 1/4 + 1/6

                   = 8 + 3 + 2 / 12

                   = 13/12 > 1.

We know that sum of all probabilities can never greater than 1.

Thus the three events A, B and C are not mutually exclusive.