14.Probability
hard

ત્રણ ઘટનાઓ $A , B$ અને $C$ ની સંભાવના અનુક્રમે $P ( A )=0.6, P ( B )=0.4$ અને $P ( C )=0.5$ આપેલ છે જો $P ( A \cup B )=0.8, P ( A \cap C )=0.3, P ( A \cap B \cap$ $C)=0.2, P(B \cap C)=\beta$ અને $P(A \cup B \cup C)=\alpha$ જ્યાં $0.85 \leq \alpha \leq 0.95,$ હોય તો $\beta$ ની કિમત ........ અંતરાલમાં રહે છે 

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

$\beta \in[0.25,0.35]$

Std 11
Mathematics

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