A,B,C are events such that
`P_(r )(A)=0.3, P_(r )(B)=0.4,P_(r )(C )=0.8`
`P_(r ) (AB)=0.08, P_(r )(AC )=0.28` and `P_(r )(B)=0.4, P_(r )(C )=0.8`
If `P_(r )(A cup B cup C) ge 0.75`, then show that `P_(r )(BC )` lies in the interval [0.23,0.48].

Answer

A

Explanation

We know that ,
`P(A) +P(B) +P(C)-P(A cap B)-P( B cup C ) -P(C cap A)+P(A cap B cap C )=P(A cup B cup C ) `
`implies 0.3+0.4+0.8-{0.8+0.28+P(BC)}+0.09=P(A cup B cup C) `
`implies 1.23 -P(BC) =P(A cup B cup C) `
where `,0.75 le P(A cup B cup B ) le 1 `
`implies 0.75 le 1.23 -P(BC) le 1 `
`implies -0.48 le -P(BC) le -0.23 `
`implies 0.23 le P(BC) le 0.48`