Q77 / 160 · TS EAMCET 2009 · Mathematics

If $A$ and $B$ are events of a random experiment such that $P(A \cup B)=\frac{4}{5}, P(\bar{A} \cup \bar{B})=\frac{7}{10}$ and $P(B)=\frac{2}{5}$, then $P(A)$ equals

  • A
    $\frac{9}{10}$
  • B
    $\frac{8}{10}$
  • Correct answer
    C
    $\frac{7}{10}$
  • D
    $\frac{3}{5}$

Solution

$\begin{aligned} & \text { Given, } P(\bar{A} \cup \bar{B})=P(\overline{A \cap B})=\frac{7}{10} \\ & \text { Since, } P(A \cap B)+P(\overline{A \cap B})=1\end{aligned}$ $\begin{array}{ll}\Rightarrow \quad P(A \cap B)=1-\frac{7}{10}=\frac{3}{10} \\ \text { Also, } P(A \cup B)=P(A)+P(B)-P(A \cap B)\end{array}$ $\begin{aligned} \Rightarrow \quad \frac{4}{5} & =P(A)+\frac{2}{5}-\frac{3}{10} \\ \Rightarrow \quad P(A) & =\frac{4}{5}-\frac{2}{5}+\frac{3}{10} \\ & =\frac{2}{5}+\frac{3}{10}=\frac{7}{10}\end{aligned}$

Hence, the correct answer is option: 'C'.

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