A) \[\frac{1}{2}\]
B) \[\frac{2}{5}\]
C) \[\frac{1}{5}\]
D) \[\frac{2}{3}\]
Correct Answer: B
Solution :
Probability of success \[=\frac{2}{6}=\frac{1}{3}=p\] Probability of failure \[=1-\frac{1}{3}=\frac{2}{3}=q\] Probability that success occurs in even number of tosses \[=P(FS)+P(FFFS)+P(FFFFFS)+..........\] \[=pq+{{q}^{3}}p+{{q}^{5}}p+........\]\[=\frac{pq}{1-{{q}^{2}}}=\frac{2}{5}.\]
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