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# Let A and B be two events such that $P ( A') = 0.3 \: P(B)=0.4 \: and \: P(A \cap B') = 0.5$ then $P(B/A\: UB')=$

$\begin {array} {1 1} (1)\;\large\frac{1}{2} & \quad (2)\;\large\frac{3}{4} \\ (3)\;\large\frac{1}{4} & \quad (4)\;\large\frac{3}{7} \end {array}$

$P ( B/A \cup B') = \large\frac{P( B \cap ( A \cup B'))}{P(A \cup B')}$
$= \large\frac{((B \cap A) \cup (B \cap B'))}{P(A)+P(B')-P(A \cap B')}$
But $P((B \cap A) \cup ( B \cap B'))$
$= P(B \cap A ) = P(A)-P( A \cap B')$
$= 0.7 - 0.5$
$= 0.2$
and $P(A)+P(B')-P(A \cap B')=0.7+0.6-0.5$
=$0.8$
$\therefore$ From (1), $( P(B) / A \cup B') = \large\frac{0.2}{0.8}$
$\large\frac{1}{4}$
Ans : (3)