Statistics And Probability Question 176
Question: If $ \frac{1+4p}{4},\frac{1-p}{2} $ and $ \frac{1-2p}{2} $ are the probabilities of three mutually exclusive events, then value of p is
Options:
A) $ \frac{1}{2} $
B) $ \frac{1}{3} $
C) $ \frac{1}{4} $
D) $ \frac{2}{3} $
Show Answer
Answer:
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} $
BETA
