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  • A, B, C are three events for which <img alt="\mathrm{P(A)=0.6, P(B)=0.4, P(C)=0.5, P(A \cup B)=0.8, P(A \cap C)=0.3}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BP%28A%29%3D0.6

A, B, C are three events for which \mathrm{P(A)=0.6, P(B)=0.4, P(C)=0.5, P(A \cup B)=0.8, P(A \cap C)=0.3} and \mathrm{P(A \cap B \cap C)=0.2}. If \mathrm{P(A \cup B \cup C) \geq 0.85} then the interval of values of \mathrm{P(B \cap C)} is

Option: 1

[0.2,0.35]


Option: 2

[0.55,0.7]


Option: 3

[0.2,0.55]

 


Option: 4

None of these


Answers (1)

best_answer

\begin{aligned} & \mathrm{P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C)} \\ \\& \mathrm{ -P(C \cap A)+P(A \cap B \cap C) }\\ \\& \mathrm{=0.6+0.4+0.5-0.2-P(B \cap C)-0.3+0.2=1.2-P(B \cap C) \text {, }} \\ & \end{aligned}

\\\text{because} \mathrm{P(A \cup B)=P(A)+P(B)-P(A \cap B) \Rightarrow 0.8=0.6+0.4-P(A \cap B).}\\ \\\text{But } \mathrm{0.85 \leq P(A \cup B \cup C) \leq 1}\\ \\\therefore \mathrm{ \quad 0.85 \leq 1.2-P(B \cap C) \leq 1 \quad \Rightarrow \quad 0.2 \leq P(B \cap C) \leq 0.35 \text {. }}

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Deependra Verma

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