A) \[\frac{1}{3}\]
B) \[\frac{2}{3}\]
C) \[\frac{2}{5}\]
D) \[\frac{3}{5}\]
Correct Answer: A
Solution :
Let\[q=1-p\]. Since, head appears first time in an even throw 2 or 4 or 6. \[\therefore \] \[\frac{2}{5}=qp+{{q}^{3}}p+{{q}^{5}}p+...\] \[\therefore \] \[\frac{2}{5}=\frac{qp}{1-{{q}^{2}}}\] \[\Rightarrow \] \[\frac{2}{5}=\frac{(1-p)p}{1-{{(1-p)}^{2}}}\] \[\frac{2}{5}=\frac{1-p}{2-p}\] \[\frac{2}{5}=\frac{1-p}{2-p}\] \[\Rightarrow \] \[4-2p=5-5p\] \[\Rightarrow \] \[3p=1\] \[\Rightarrow \]\[p=\frac{1}{3}\]You need to login to perform this action.
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