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

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