| List-I | List-II |
| (A) \[P({{E}_{2}})\] | (i) 1/4 |
| (B) \[P({{E}_{1}}\cup {{E}_{2}})\] | (ii) 5/8 |
| (C) \[P({{\bar{E}}_{1}}/{{\bar{E}}_{2}})\] | (iii) 1/8 |
| (D)\[P({{E}_{1}}/{{\bar{E}}_{2}})\] | (iv) 1/2 |
| (v) 3/8 | |
| (vi) 3/4 |
A) A-(ii) B-(iii) C-(vi) D-(i)
B) A-(iv) B-(v) C-(vi) D-(i)
C) A-(iv) B-(ii) C-(vi) D-(i)
D) A-(i) B-(ii) C-(iii) D-(iv)
Correct Answer: C
Solution :
(A) Given,\[P({{E}_{1}})=\frac{1}{4},\,\,P\left( \frac{{{E}_{1}}}{{{E}_{2}}} \right)=\frac{1}{4}\] and \[P\left( \frac{{{E}_{2}}}{{{E}_{1}}} \right)=\frac{1}{2}\] \[\Rightarrow \] \[\frac{P({{E}_{2}}\cap {{E}_{1}})}{P({{E}_{1}})}=\frac{1}{2}\] \[\Rightarrow \] \[P({{E}_{2}}\cap {{E}_{1}})=\frac{1}{8}\] Also, \[P\left( \frac{{{E}_{1}}}{{{E}_{2}}} \right)=\frac{1}{4}\] \[\Rightarrow \] \[\frac{P({{E}_{1}}\cap {{E}_{2}})}{P({{E}_{2}})}=\frac{1}{4}\] \[\Rightarrow \] \[\frac{1}{8P({{E}_{2}})}=\frac{1}{4}\] \[\Rightarrow \] \[P({{E}_{2}})=\frac{1}{2}\] (B)\[P({{E}_{1}}\cup {{E}_{2}})=P({{E}_{1}})+P({{E}_{2}})-P({{E}_{1}}\cap {{E}_{2}})\] \[=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}=\frac{5}{8}\] (C)\[P\left( \frac{{{{\bar{E}}}_{1}}}{{{{\bar{E}}}_{2}}} \right)=\frac{P({{{\bar{E}}}_{1}}\cap {{{\bar{E}}}_{2}})}{P({{{\bar{E}}}_{2}})}\] \[=\frac{1-P({{E}_{1}}\cup {{E}_{2}})}{1-P({{E}_{2}})}=\frac{1-\frac{5}{8}}{1-\frac{1}{2}}\] \[=\frac{3}{4}\] (D)\[P\left( \frac{{{E}_{1}}}{{{E}_{2}}} \right)=\frac{P({{E}_{1}}\cap {{{\bar{E}}}_{2}})}{1-P({{\overline{E}}_{2}})}\] \[=\frac{\frac{1}{4}-\frac{1}{8}}{1-\frac{1}{2}}=\frac{\frac{1}{8}}{\frac{1}{2}}=\frac{1}{4}\]
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