A) \[\frac{1}{3}\le p\le \frac{1}{2}\]
B) \[\frac{1}{3}<p<\frac{1}{2}\]
C) \[\frac{1}{2}\le p\le \frac{2}{3}\]
D) \[\frac{1}{2}<p<\frac{2}{3}\]
Correct Answer: A
Solution :
Since \[\frac{(1+3p)}{3},\,\,\frac{(1-p)}{4}\] and \[\left( \frac{1-2p}{2} \right)\] are the probabilities of the three events, we must have \[0\le \frac{1+3p}{3}\le 1,\,\,0\le \frac{1-p}{4}\le 1\] and \[0\le \frac{1-2p}{2}\le 1\] \[\Rightarrow -1\le 3p\le 2,\,\,-3\le p\le 1\] and \[-1\le 2p\le 1\] \[\Rightarrow -\frac{1}{3}\le p\le \frac{2}{3},\,\,-3\le p\le 1\] and \[-\frac{1}{2}\le p\le \frac{1}{2}\] Also as \[\frac{1+3p}{3},\,\,\frac{1-p}{4}\] and \[\frac{1-2p}{2}\] are the probabilities of three mutually exclusive events \[0\le \frac{1+3p}{3}+\frac{1-p}{4}+\frac{1-2p}{2}\le 1\] \[\Rightarrow 0\le 4+12p+3-3p+6-12p\le 12\]\[\Rightarrow \frac{1}{3}\le p\le \frac{13}{3}\] Thus the required value of \[p\] are such that Max.\[\left\{ -\frac{1}{3},\,\,-3,\,\,-\frac{1}{2},\,\,\frac{1}{3} \right\}\le p\] min.\[\left\{ \frac{2}{3},\,\,1,\,\,\frac{1}{2},\,\,\frac{13}{3} \right\}\] \[\Rightarrow \frac{1}{3}\le p\le \frac{1}{2}.\]
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