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જો $A$ અને $B$ એવી ઘટનાઓ છે કે જેથી $P(A\, \cup \,\,B)\,\, = \,\,\frac{3}{4},\,\,P(A\,\, \cap \,\,B)\,\, = \,\,\frac{1}{4}\,,\,P(\overline A )\,\, = \,\,\frac{2}{3},\,$ હોય , તો $P(\overline A \,\, \cap \,\,B)\,$ બરાબર શું થાય?
$5/12$
$3/8$
$5/8$
$1/4$
Solution
$\begin{gathered}
P(A\, \cup \,\,B)\,\, = \,\,\frac{3}{4},\,\,P(A\, \cap \,\,B)\,\, = \,\,\frac{1}{4} \hfill \\
P(\overline A )\,\, = \,\,\frac{2}{3}\,\,\,\, \Rightarrow \,P(A)\,\, = \,\,\frac{1}{3} \hfill \\
\therefore \,\,P(A\, \cap \,\,B)\,\, = \,\,P(A)\,\, + \,\,P(B)\,\, – \,\,P(A\, \cup \,\,B) \hfill \\
\frac{1}{4}\,\, = \,\,\frac{1}{3}\,\, + \,\,P(B)\,\, – \,\,\frac{3}{4}\,\, \Rightarrow \,\,P(B)\,\, = \,\,\frac{2}{3} \hfill \\
P(\overline A \, \cap \,\,B)\,\, = \,\,P(B)\,\, – \,\,P(A\,\, \cap \,\,B)\,\,\,\,\,\, \hfill \\
= \,\,\frac{2}{3}\,\,\, – \,\,\frac{1}{4}\,\, = \,\,\frac{{8\,\, – \,\,3}}{{12}}\,\, = \,\,\frac{5}{{12}}. \hfill \\
\end{gathered} $
