A) \[\frac{3}{5}\]
B) \[\frac{5}{6}\]
C) \[\frac{1}{3}\]
D) \[\frac{2}{3}\]
Correct Answer: B
Solution :
Consider, the events \[{{E}_{1}},\,{{E}_{2}}\] and A \[P({{E}_{1}})=\frac{1}{2}.\,P({{E}_{2}})=\frac{1}{2}\] \[P({{E}_{1}}/A)=\frac{20\times 100}{100\times 100},\,\,P({{E}_{2}}/A)=\frac{20\times 20}{100\times 100}\] \[\therefore \] Required probability \[P(A/{{E}_{1}})=\frac{P({{E}_{1}})\times P({{E}_{1}}/A)}{P({{E}_{1}})\times P({{E}_{1}}/A)+P({{E}_{2}})\times P({{E}_{2}}/A)}\] \[=\frac{\frac{1}{2}\times \frac{20\times 100}{100\times 100}}{\frac{1}{2}\times \frac{20\times 100}{100\times 100}+\frac{1}{2}\times \frac{20\times 20}{100\times 100}}\] \[=\frac{20\times 100}{20\times (100+20)}=\frac{100}{120}=\frac{5}{6}\]
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