Question

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

• A. 1
• B. $3 / 4$
• C. $1 / 4$
• D. None of these

Answer 1

Answer: (B)

$P(E)+P\left(E^{\prime}\right)=1=1+\lambda+\lambda^{2}+(1+\lambda)^{2}$
or $2 \lambda^{2}+3 \lambda+1=0$
or $(2 \lambda+1)(\lambda+1)=0$
or $\lambda=-1,-\frac{1}{2}$
Then, $P(E)=1+(-1)+(-1)^{2}=1$ (not possible $)$
$=1+\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right)^{2}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$