The probabilities of three events `A ,B ,a n dC` are `P(A)=0. 6 ,P(B)=0. 4 ,a n dP(C)=0. 5.` If `P(AuuB)=0. 8 ,P(AnnC)=0. 3 ,P(AnnBnnC)=0. 2 ,a n dP(A

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The probabilities of three events `A ,B ,a n dC` are `P(A)=0. 6 ,P(B)=0. 4 ,a n dP(C)=0. 5.` If `P(AuuB)=0. 8 ,P(AnnC)=0. 3 ,P(AnnBnnC)=0. 2 ,a n dP(A

The probabilities of three events `A ,B ,a n dC` are `P(A)=0. 6 ,P(B)=0. 4 ,a n dP(C)=0. 5.` If `P(AuuB)=0. 8 ,P(AnnC)=0. 3 ,P(AnnBnnC)=0. 2 ,a n dP(AuuBuuC)geq0. 85 ,` then find the range of `P(BuuC)dot`

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

We have,
`P(A nn B) = P(A) + P(B) - P(A uu B) = 0.6 + 0.4 - 0.8 = 0.2`
Also, `P(A uu B uu C) = P(A) + P(B) + P(C) - P(A nn C) + P(A nn B nn C) - P(A nn B) - P(B nn C)`
`implies P(B nn C) = 1.2 - P(A uu B uu C)" "(1)`
Now, `0.85 le P(A uu B uu C) le 1`
From Eq. (1) `0.2 le P(B nnC) le 0.35`

5 views Answered 2023-02-05 15:29