A) \[1/3\]
B) \[2/3\]
C) \[2/5\]
D) \[3/5\]
Correct Answer: A
Solution :
Let p be the probability of getting a head, then \[q=1-p\]. Since, head appears first time in an even throw i.e. 2 or 4 or 6... . \[\therefore \] \[\frac{2}{5}=ap+{{q}^{3}}p+{{q}^{5}}p+...\] \[\Rightarrow \] \[\frac{2}{5}=\frac{ap}{1-{{q}^{2}}}\] \[\Rightarrow \] \[\frac{2}{5}=\frac{(1-p)p}{1-{{(1-p)}^{2}}}\] \[(\because q=1-p)\] \[\Rightarrow \] \[\frac{2}{5}=\frac{1-p}{2-p}\] \[\Rightarrow \] \[4-2p=5-5p\Rightarrow 3p=1\] \[\Rightarrow \] \[p=\frac{1}{3}\]
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