A) \[0.1\]
B) \[0.3\]
C) \[0.4\]
D) \[0.2\]
Correct Answer: D
Solution :
Given, \[P({{\bar{E}}_{2}})=0.6=P({{E}_{1}}\cup {{E}_{2}})\] Since, \[{{E}_{1}},\,\,{{E}_{2}}\] are mutually exclusive events, then \[P({{E}_{1}}\cap {{E}_{2}})=0\] Now,\[P({{E}_{1}}\cup {{E}_{2}})=P({{E}_{1}})+P({{E}_{2}})-P({{E}_{1}}\cap {{E}_{2}})\] \[\Rightarrow \] \[0.6\,=\,P({{E}_{1}})+1-P({{\bar{E}}_{2}})-0\] \[\Rightarrow \] \[0.6=P({{E}_{1}})+0.4\] \[\Rightarrow \] \[P({{E}_{1}})=0.2\]
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