A) 5/8
B) 1/2
C) 1/3
D) 2/3
Correct Answer: D
Solution :
The chance of head \[=\frac{1}{2}\] and not of head \[=\frac{1}{2}\] Since \[A\] has first throw, he can win in the first, third, ? \[\therefore \]Probability of A?s winning \[=\frac{1}{2}+{{\left( \frac{1}{2} \right)}^{2}}.\frac{1}{2}+{{\left( \frac{1}{2} \right)}^{4}}.\frac{1}{2}+........\] \[=\frac{1}{2}+{{\left( \frac{1}{2} \right)}^{3}}+{{\left( \frac{1}{2} \right)}^{5}}+.........=\frac{2}{3}\].
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