Q110 / 160 · AP EAMCET 2014 · Mathematics

If $A, B$ and $C$ are mutually exclusive and exhaustive events of a random experiment such that $P(B)=\frac{3}{2} P(A)$ and $P(C)=\frac{1}{2} P(B)$, then $P(A \cup C)$ equals to

  • A
    $\frac{10}{13}$
  • B
    $\frac{3}{13}$
  • C
    $\frac{6}{13}$
  • Correct answer
    D
    $\frac{7}{13}$

Solution

Given, $\quad P(B)=\frac{3}{2} P(A)$ and $$ P(C)=\frac{1}{2} P(B) $$ Since, $A, B$ and $C$ are exclusive events. $$ \begin{aligned} & \therefore & P(A)+P(B)+P(C) & =1 \\ & \therefore & P(A)+\frac{3}{2} P(A)+\frac{1}{2} \times \frac{3}{2} P(A) & =1 \\ & \Rightarrow & P(A)\left(1+\frac{3}{2}+\frac{3}{4}\right) & =1 \\ & & \frac{13}{4} P(A) & =1 \\ & & P(A) & =\frac{4}{13} \\ & \therefore & P(C)=\frac{1}{2} \times \frac{3}{2} P(A) & =\frac{3}{4} \times \frac{4}{13}=\frac{3}{13} \end{aligned} $$ Also, $A, B$ and $C$ are mutually exclusive. $$ \begin{aligned} & \therefore P(A \cap B)=P(B \cap C)=P(C \cap A)=0 \\ & \therefore P(A \cup C)=P(A)+P(B)-0 \\ & =\frac{4}{13}+\frac{3}{13}=\frac{7}{13} \end{aligned} $$

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

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