Let A and B be two events such that Poverline | vedclass

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Question

(a) $P\left( {\overline {A \cup B} } \right) = \frac{1}{6};\,\,P\left( {A \cap B} \right) = \frac{1}{4}$,

$P\left( {\bar A} \right) = \frac{1}{4} \

Vedclass · Accepted Answer

Independent but not equally likely

Vedclass · Answer

(a) $P\left( {\overline {A \cup B} } ight) = \frac{1}{6};\,\,P\left( {A \cap B} ight) = \frac{1}{4}$,

$P\left( {\bar A} ight) = \frac{1}{4} \Rightarrow P\left( A ight) = \frac{3}{4}$,

$P\left( {\overline {A \cup B} } ight) = 1 - P\left( {A \cup B} ight) = 1 - P\left( A ight) - P\left( B ight) + P\left( {A \cap B} ight)$

$&rArr;$ $\frac{1}{6} = \frac{1}{4} - P\left( B ight) + \frac{1}{4}$ $&rArr;$ $P\left( B ight) = \frac{1}{3}$.

Since $P\left( {A \cap B} ight) = P\left( A ight)P\left( B ight)$ and $P\left( A ight) 
e P\left( B ight)$

$	herefore$ $ A$ and $B$ are independent but not equally likely.

Let $A$ and $B$ be two events such that $P\overline {(A \cup B)} = \frac{1}{6},P(A \cap B) = \frac{1}{4}$ and $P(\bar A) = \frac{1}{4},$ where $\bar A$ stands for complement of event $A$. Then events $A$ and $B$ are

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