A) \[\frac{1}{3}\]
B) \[\frac{2}{3}\]
C) \[\frac{1}{4}\]
D) \[\frac{3}{4}\]
Correct Answer: B
Solution :
Probability of getting the head by A in first chance \[=\frac{1}{2}\] Probability of losing the toss by A in first chance \[=1-\frac{1}{2}=\frac{1}{2}\] Probability of getting head by B in first chance \[=\left( \frac{1}{2} \right)\times \left( \frac{1}{2} \right)={{\left( \frac{1}{2} \right)}^{2}}\] Probability of getting head by A in second chance \[={{\left( \frac{1}{2} \right)}^{2}}\times \frac{1}{2}={{\left( \frac{1}{2} \right)}^{3}}\] ???????????.. ???????????.. The probability that A wins the toss. \[=\left( \frac{1}{2} \right)+{{\left( \frac{1}{2} \right)}^{3}}+{{\left( \frac{1}{2} \right)}^{5}}+....=\frac{1/2}{1-{{\left( \frac{1}{2} \right)}^{2}}}=\frac{2}{3}\]
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