Let `A ,B ,C` be three events such that `P(A)=0. 3 ,P(B)=0. 4 ,P(C)=0. 8 ,P(AnnB)=0. 88 ,P(AnnC)=0. 28 ,P(AnnBnnC)=0. 09.` If `P(AuuBuuC)geq0. 75 ,` then show that `0. 23lt=P(BnnC)lt=0. 48.`

Explanation

Given` P(AuuBuuC)≥0.75`


and ,
`P(A)=0. 3 ,P(B)=0. 4 ,P(C)=0. 8 `
`P(AnnB)=0. 88 ,P(AnnC)=0. 28
,P(AnnBnnC)=0. 09.`

So,


`=>0.75leP(AuuBuuC)le1`

`=>0.75leP(A)+P(B)+P(C)−P(AnnB)−P(BnnC)−P(AnnC)+P(AnnBnnC)le1 `
`=>0.75le0.3+0.4+0.8−0.08−P(BnnC)−0.28+0.09le1 `

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