A) \[\frac{1}{2}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{4}\]
D) \[\frac{2}{3}\]
Correct Answer: A
Solution :
[a] \[\frac{1+4p}{4},\frac{1-p}{2},\frac{1-2p}{2}\] are probabilities of the three mutually exclusive events, then \[0\le \frac{1+4p}{4}\le 1,0\le \frac{1-p}{2}\le 1,0\le \frac{1-2p}{2}\le 1\] and \[0\le \frac{1+4p}{4}+\frac{1-p}{2}+\frac{1-2p}{2}\le 1\] \[\therefore -\frac{1}{4}\le p\le \frac{3}{4},-1\le p\le 1,-\frac{1}{2}\le p\frac{1}{2},\frac{1}{2}\le p\le \frac{5}{2}\]\[\therefore \frac{1}{2}\le p\le \frac{1}{2}\] [The intersection of above four intervals] \[\therefore \,\,\,p=\frac{1}{2}\]
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