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Question

# If A and B are two events such that $P\left(A\cap B\right)=\frac{1}{3},P\left(A\cup B\right)=\frac{5}{6}\mathrm{and}P\left(\overline{)\mathit{B}}\right)=\frac{1}{3}$, then P(A)=_____________.

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Solution

## For two events A and B given P(A⋂B) = $\frac{1}{3}$ P(A⋃B) = $\frac{5}{6}$ $P\left(\overline{)B}\right)=\frac{1}{3}$ $\mathrm{i}.\mathrm{e}P\left(B\right)=1-P\left(\overline{)B}\right)\phantom{\rule{0ex}{0ex}}=1-\frac{1}{3}\phantom{\rule{0ex}{0ex}}P\left(B\right)=\frac{2}{3}$ using P(A⋃B) = P(A) + P(B) – P(A⋂B) $\begin{array}{rcl}\frac{5}{6}& =& P\left(A\right)+\frac{2}{3}-\frac{1}{3}\\ \mathrm{i}.\mathrm{e}P\left(A\right)& =& \frac{5}{6}-\frac{2}{3}+\frac{1}{3}\\ & =& \frac{5}{6}-\frac{1}{3}\\ & =& \frac{5-2}{6}\\ & =& \frac{3}{6}\\ \mathrm{i}.\mathrm{e}P\left(A\right)& =& \frac{1}{2}\end{array}$

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