Question

Let $E$ be an event which is neither a certainty nor an impossibility. If probability is such that $P\left(E\right)=1+\lambda +{\lambda }^2$ and $P\left(E^{\prime }\right)=(1+\lambda {)}^2$ in terms of an unknown $\lambda ,$ then the number of possible values of $\lambda$ is

Answer 1

$P\left(E\right)+P\left(E^{\prime }\right)=1$
$\Rightarrow 1+\lambda +{\lambda }^2+(1+\lambda {)}^2=1$
$\Rightarrow 2{\lambda }^2+3\lambda +1=0$
$\Rightarrow (2\lambda +1)(\lambda +1)=0$
$\Rightarrow \lambda =-\frac{1}{2},-1$
When $\lambda =-1$
$P\left(E\right)=1-1+1=1$
Hence $\lambda \ne -1$
$(\because P(E\left)\text{not certain}\right)$
When $\lambda =-\frac{1}{2}$
$\Rightarrow P\left(E\right)=\frac{3}{4}$
$\therefore$ Only one value of $\lambda$ exists.