Question

23. The probabilities of three events A, B. Care such that
P(A)=0.3,P(B)=04.P(C)=08 P(AnB)=008. P(ANC)=0.28, P(AnBnC)=0,09
and P(AUBUC)20.75, show that P(BOC) lies in the internal (0.23.0.48]​

Answer 1

Answer:

the probabilities of three events

A,B,C

are such that :

P(A)= 0.3 P(B) 0.4 P(C) 0.8

P(BΠC)

lies in the interval

[0.21, 0.46]

its the answer......