Question

If $\frac{1+3p}{3},\frac{1-2p}{2}$ are probabilities of two mutually exclusive event, then $p$ lies in the interval

• A. $[-\frac{1}{3},\frac{1}{2}]$
• B. $(-\frac{1}{2},\frac{1}{2})$
• C. $[-\frac{1}{2},\frac{2}{3}]$
• D. $(-\frac{1}{3},\frac{2}{3})$

Answer 1

The correct option is A $[-\frac{1}{3},\frac{1}{2}]$

$Probabilitiesare\frac{1+3p}{3}and\frac{1-2p}{2}so,0\le \frac{1+3p}{3}\le 1\Rightarrow 0\le 1+3p\le 3\Rightarrow -1/3\le p\le 2/3Alse,0\le \frac{1-2p}{2}\le 1\Rightarrow 0\le 1-2p\le 2\Rightarrow -1\le -2p\le 1\Rightarrow -1\le 2p\le 1\Rightarrow -1/2\le p\le 1/2Intersectionrangeofpforbothprobabilitiestobevalid=-1/3\le p\le 1/2$