If A,B and C are three events, such that P(A)=0.3, P(B)=0.4, P(C)=0.8, P(AB)=0.08, P(AC)=0.28, P(ABC)=0.09. If `P(AcupBcupC)ge0.75`, then show that P(BC) kies in the interval `0.23lexle0.48`

Options

A. `P(B cap C) le 0.23`
B. `P(B cap C) le 0.48`
C. `0.23 le P(B cap C) le 0.48`
D. `0.23 le P(B cap C) le 0.48`

Answer

C,D

Explanation

We have that the probability of occurrence of an event is always less than or equal to 1 and it is given that `P(A cup B cup C) ge 0.75`.
`therefore 0.75 le P(A cup B cup C) le 1`
`implies 0.75 le P(A)+P(B)+P(C )-P(A cap B)-P(B cap C)-P(A cap C)+P(A cap B cap C) le 1`
`implies 0.75 le 0.3+0.4+0.8-0.08-P(B cap C)-0.28+0.09 le 1`
`implies 0.75 le 1.59 -0.36-P(B cap C) le 1`
`implies 0.75 le 1.23-P(B cap C) le 1`
`implies -0.48 le -P(B cap C) le -0.23 implies 0.23 le P(B cap C) le 0.48`