14.Probability
hard

तीन घटनाओं $A , B$ तथा $C$ की प्रायिकताएं $P ( A )=0.6$, $P ( B )=0.4$ तथा $P ( C )=0.5$ है। यदि $P ( A \cup B )=0.8$, $P ( A \cap C )=0.3, P ( A \cap B \cap C )=0.2, P ( B \cap$ $C )=\beta$ तथा $P ( A \cup B \cup C )=\alpha$, जहाँ $0.85 \leq \alpha \leq 0.95$, तो $\beta$ निम्न में से किस अंतराल में है 

A

$[0.36,0.40]$

B

$[0.35,0.36]$

C

$[0.25,0.35]$

D

$[0.20,0.25]$

(JEE MAIN-2020)

Solution

$P ( A \cup B )= P ( A )+ P ( B )- P ( A \cap B )$

$0.8=0.6+0.4- P ( A \cap B )$

$P ( A \cap B )=0.2$

$P ( A \cup B \cup C )=\Sigma P ( A )-\Sigma P ( A \cap B )+ P ( A \cap B \cap C )$

$\alpha=1.5-(0.2+0.3+\beta)+0.2$

$\alpha=1.2-\beta \in[0.85,0.95]$

(where $\alpha \in[0.85,0.95])$

$\beta \in[0.25,0.35]$

Standard 11
Mathematics

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