A) \[\frac{1}{4}and\frac{3}{4}\]
B) \[\frac{1}{2}and\frac{1}{2}\]
C) \[\frac{2}{3}and\frac{1}{3}\]
D) \[\frac{1}{5}and\frac{4}{5}\]
E) 0 and 1
Correct Answer: C
Solution :
The probability of winning A \[=\frac{1}{2}+\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}+....\] \[=\frac{1}{2}+{{\left( \frac{1}{2} \right)}^{3}}+{{\left( \frac{1}{2} \right)}^{5}}+....\] \[=\frac{1}{2}\left[ \frac{1}{1-\frac{1}{4}} \right]=\frac{2}{3}\] The probability of winning B \[=\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}+.....\] \[={{\left( \frac{1}{2} \right)}^{2}}+{{\left( \frac{1}{2} \right)}^{4}}+.....\] \[=\frac{1}{4}\left[ \frac{1}{1-\frac{1}{4}} \right]=\frac{1}{3}\]
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