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If $A$ and $B$ are two mutually exclusive and exhaustive events
with $P(B)=3 P(A)$, then what is the value of $P(\bar{B})$ ?
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with $P(B)=3 P(A)$, then what is the value of $P(\bar{B})$ ?
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The correct answer is:
$1 / 4$
A and B are mutually exclusive and exhaustive events with $P(A \cap B)=0, P(A \cup B)=1$
we know that $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\Rightarrow 1=P(A)+3 P(A)$
$\Rightarrow \quad P(A)=\frac{1}{4}$
$\therefore P(B)=\frac{3}{4}$
Hence, $P(\bar{B})=1-P(B)=1-\frac{3}{4}=\frac{1}{4}$
we know that $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\Rightarrow 1=P(A)+3 P(A)$
$\Rightarrow \quad P(A)=\frac{1}{4}$
$\therefore P(B)=\frac{3}{4}$
Hence, $P(\bar{B})=1-P(B)=1-\frac{3}{4}=\frac{1}{4}$
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