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Let A ,B ,C be three events such that P(...

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.`

A

`P(B nn C)le 0.23`

B

`P(B nn C) le 0.48`

C

`0.23 le P(B nn C)le 0.48`

D

`0.23 le P(B nn C) ge 0.48`

Text Solution

Verified by Experts

The correct Answer is:
C
Given, `P(A)=0.3, P(B)=0.4, P(C )=0.08`,
`P(A nn B)=0.08, P(A nn C) = 0.28`,
`P(A nn B nn C) = 0.09`
Since, `P(A uu B uu C) ge 0.75`
`rArr P(A)+P(B)+P(C )-P(A nn B)-P(A nn C)-P(B nn C)+P(A nn B nnC)ge 0.75`
`rArr 0.3+0.4+0.8-0.08-0.28-P(B nn C)+0.009 ge 0.75`
`rArr P(B nn C) le 0.48`
Also, `P(A uu B uu C) le 1`
`rArr 1.23-P(B nn C) le 1`
`rArr P(B nn C) ge 0.23`
`therefore 0.23 le P(B nn C) le 0.48`
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