Question

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

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

Answer 1

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

For both the given probabilities to be mutually exclusive

$0\le \frac{1+3p}{3}\le 1,$ and $0\le \frac{1-2p}{2}\le 1$

$\Rightarrow 0\le 1+3p\le 3$ and $0\le 1-2p\le 2$

$\Rightarrow -1\le 3p\le 2$ and $-1\le -2p\le 1$

$\Rightarrow -\frac{1}{3}\le p\le \frac{2}{3}$ and $-\frac{1}{2}\le p\le \frac{1}{2}$

$\therefore p\in [-\frac{1}{3},\frac{1}{2}]$