Let A and B be two events such that P(overlin | vedclass

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(A) Given $P(\overline{A \cup B}) = \frac{1}{6}$ and $P(A \cap B) = \frac{1}{4}$.Since $P(\bar{A}) = \frac{1}{4}$,we have $P(A) = 1 - \frac{1}{4} = \f

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Independent but not equally likely

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(A) Given $P(\overline{A \cup B}) = \frac{1}{6}$ and $P(A \cap B) = \frac{1}{4}$.Since $P(\bar{A}) = \frac{1}{4}$,we have $P(A) = 1 - \frac{1}{4} = \frac{3}{4}$.We know that $P(\overline{A \cup B}) = 1 - P(A \cup B) = 1 - [P(A) + P(B) - P(A \cap B)]$.Substituting the values: $\frac{1}{6} = 1 - [\frac{3}{4} + P(B) - \frac{1}{4}] = 1 - [\frac{1}{2} + P(B)] = \frac{1}{2} - P(B)$.Thus,$P(B) = \frac{1}{2} - \frac{1}{6} = \frac{3-1}{6} = \frac{2}{6} = \frac{1}{3}$.Now,check for independence: $P(A) 	imes P(B) = \frac{3}{4} 	imes \frac{1}{3} = \frac{1}{4}$.Since $P(A \cap B) = P(A) 	imes P(B) = \frac{1}{4}$,the events $A$ and $B$ are independent.Since $P(A) = \frac{3}{4}$ and $P(B) = \frac{1}{3}$,$P(A) 
eq P(B)$,so they are not equally likely.Therefore,the events $A$ and $B$ are independent but not equally likely.

Box	White	Black
$B_1$	$1$	$2$
$B_2$	$3$	$1$
$B_3$	$2$	$3$

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 the complement of event $A$. Then events $A$ and $B$ are

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