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14.Probability
hard
The probabilities of three events $A , B$ and $C$ are given by $P ( A )=0.6, P ( B )=0.4$ and $P ( C )=0.5$ If $P ( A \cup B )=0.8, P ( A \cap C )=0.3, P ( A \cap B \cap$ $C)=0.2, P(B \cap C)=\beta$ and $P(A \cup B \cup C)=\alpha$ where $0.85 \leq \alpha \leq 0.95,$ then $\beta$ lies in the interval
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
