Q79 / 160 · AP EAMCET 2009 · Mathematics

Suppose that $E_1$ and $E_2$ are two events of a random experiment such that $P\left(E_1\right)=\frac{1}{4}$, $P\left(E_2 / E_1\right)=\frac{1}{2}$ and $P\left(E_1 / E_2\right)=\frac{1}{4}$, observe the lists given below Figure
The correct matching of the List I from the List II is (A) (B) (C) (D)

  • A
    (ii) (iii) (vi) (i)
  • B
    (iv) (v) (vi) (i)
  • Correct answer
    C
    (iv) (ii) (vi) (i)
  • D
    (i) (ii) (iii) (iv)

Solution

(A) Given, $P\left(E_1\right)=\frac{1}{4}, P\left(\frac{E_1}{E_2}\right)=\frac{1}{4}$ and $P\left(\frac{E_2}{E_1}\right)=\frac{1}{2}$ $\begin{array}{ll}\Rightarrow & \frac{P\left(E_2 \cap E_1\right)}{P\left(E_1\right)}=\frac{1}{2} \\ \Rightarrow & P\left(E_2 \cap E_1\right)=\frac{1}{8} \\ \text { Also, } & P\left(\frac{E_1}{E_2}\right)=\frac{1}{4} \\ \Rightarrow & \frac{P\left(E_1 \cap E_2\right)}{P\left(E_2\right)}=\frac{1}{4} \\ \Rightarrow & \frac{1}{8 P\left(E_2\right)}=\frac{1}{4} \\ \Rightarrow & P\left(E_2\right)=\frac{1}{2}\end{array}$ $\begin{aligned} (B) \quad P\left(E_1 \cup E_2\right) & =P\left(E_1\right)+P\left(E_2\right)-P\left(E_1 \cap E_2\right) \\ & =\frac{1}{4}+\frac{1}{2}-\frac{1}{8}=\frac{5}{8} \end{aligned}$ $\begin{aligned} (C) \quad P\left(\frac{\bar{E}_1}{\bar{E}_2}\right) & =\frac{P\left(\bar{E}_1 \cap \bar{E}_2\right)}{P\left(\bar{E}_2\right)} \\ & =\frac{1-P\left(E_1 \cup E_2\right)}{1-P\left(E_2\right)}=\frac{1-\frac{5}{8}}{1-\frac{1}{2}} \\ & =\frac{3}{4} \end{aligned}$ $\begin{aligned} (D) \quad P\left(\frac{E_1}{\bar{E}_2}\right) & =\frac{P\left(E_1 \cap \bar{E}_2\right)}{P\left(\bar{E}_2\right)} \\ & =\frac{P\left(E_1\right)-P\left(E_1 \cap E_2\right)}{1-P\left(E_2\right)} \\ & =\frac{\frac{1}{4}-\frac{1}{8}}{1-\frac{1}{2}}=\frac{\frac{1}{8}}{\frac{1}{2}}=\frac{1}{4} \end{aligned}$

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

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