A) \[\frac{1}{3}\le p\le \frac{1}{2}\]
B) \[\frac{1}{2}\le p\le \frac{2}{3}\]
C) \[\frac{1}{6}\le p\le \frac{1}{2}\]
D) None of these
Correct Answer: D
Solution :
Since,\[\frac{1+4p}{p},\,\,\frac{1-p}{2}\]and\[\frac{1-2p}{2}\]are probabilities of three mutually exclusive events, therefore, \[0\le \frac{1+4p}{p}\le 1,\,\,0\le \frac{1-p}{2}\le 0,\,\,\frac{1-2p}{2}\le 1\] and \[0\le \frac{1+4p}{p}+\frac{1-p}{2}+\frac{1-2p}{2}\le 1\] \[\Rightarrow \] \[-\frac{1}{4}\le p\le \frac{3}{4},\,\,-1\le p\le 1\] \[-\frac{1}{2}\le p\le \frac{1}{2}\]and\[\frac{1}{2}\le p\le \frac{5}{2}\] \[\Rightarrow \max \] \[\left\{ \frac{-1}{4},\,\,-1,\,\,\frac{-1}{2},\,\,\frac{1}{2} \right\}\le p\le \min \left\{ \frac{3}{4},\,\,1,\,\,\frac{1}{2},\,\,\frac{5}{2} \right\}\] \[\Rightarrow \] \[\frac{1}{2}\le p\le \frac{1}{2}\Rightarrow p=\frac{1}{2}\]
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