Question

If $\frac{1+4p}{p},\frac{1-p}{4},\frac{1-2p}{2}$ are probabilities of three mutually exclusive events, then

• A. $p=\varphi$
• B. $-\frac{1}{5}\le p\le \frac{1}{2}$
• C. $-\frac{1}{2}\le p\le \frac{1}{2}$
• D. $-\frac{1}{5}\le p\le 4$

Answer 1

The correct option is A $p=\varphi$

$0\le \frac{1+4p}{p}\le 1$
$\Rightarrow -\frac{1}{3}\le p\le -\frac{1}{4}\cdots \left(i\right)$

$0\le \frac{1-p}{4}\le 1$
$\Rightarrow -3\le p\le 1\cdots \left(ii\right)$

$0\le \frac{1-2p}{2}\le 1$
$\Rightarrow -\frac{1}{2}\le p\le \frac{1}{2}\cdots \left(iii\right)$

As all three mutually exclusive events,
$\therefore 0\le \frac{1+4p}{p}+\frac{1-p}{4}+\frac{1-2p}{2}\le 1$
$\Rightarrow 0\le \frac{-5p^2+19p+4}{4p}\le 1$
$\Rightarrow -5p^2+19p+4\le 0$
$\Rightarrow 5p^2-19p-4\ge 0$
$\Rightarrow (5p+1)(p-4)\ge 0$
$\therefore p\in (-\infty ,-\frac{1}{5}]\cdots \left(iv\right)$

$-5p^2+19p+4\ge 4p$
$\Rightarrow -5p^2+15p+4\ge 0$
$\Rightarrow 5p^2-15p-4\le 0$
$\Rightarrow p\in [\frac{15-\sqrt{305}}{10},\frac{15+\sqrt{305}}{10}]\cdots \left(v\right)$

$\therefore$ From $\left(i\right),\left(ii\right),\left(iii\right),\left(iv\right)$ and $\left(v\right),$ we get
$p=\varphi$