A) \[\frac{3}{4}\]
B) \[\frac{1}{2}\]
C) \[\frac{2}{3}\]
D) \[\frac{1}{3}\]
Correct Answer: A
Solution :
\[P({{E}_{1}})=\frac{1}{2},\,\,P({{E}_{2}})=\frac{1}{3}\]and\[P({{E}_{3}})=\frac{1}{4}\] \[P({{E}_{1}}\cup {{E}_{2}}\cup {{E}_{3}})=1-P({{\bar{E}}_{1}})P({{\bar{E}}_{2}})P({{\bar{E}}_{3}})\] \[=1-\left( 1-\frac{1}{2} \right)\left( 1-\frac{1}{3} \right)\left( 1-\frac{1}{4} \right)\] \[=1-\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}=\frac{3}{4}\] Note: If the events \[{{E}_{1}},\,\,\,{{E}_{2}}\] and \[{{E}_{3}}\] are independent, then \[{{\bar{E}}_{1}},\,\,\,{{\bar{E}}_{2}}\] and \[{{\bar{E}}_{3}}\] are also independent.
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