NIMCET 2015 Question Paper with solutions

Concept:
The probability of the complement of an event is one minus the probability of the event. $\mathit{P}\left({\mathit{B}}^{\prime }\right)$ $=1-\mathit{P}\left(\mathit{B}\right)\mathit{P}(\left(\mathit{B}\cap {\mathit{C}}^{\prime }\mid \mathit{E}\right)$ $=\frac{\mathit{P}\left(\mathit{B}\right)\mathit{P}\left({\mathit{C}}^{\prime }\right)}{\mathit{P}\left(\mathit{E}\right)}$
Given, $\mathit{A}$ is targeting $\mathit{B},\mathit{B}$ and $\mathit{C}$ are targeting $\mathit{A}$. $\mathit{P}\left(\mathit{A}\right)$ : Probability of hitting the target by $\mathit{A}=\frac{2}{3}$ $\mathit{P}\left(\mathit{B}\right)$ : Probability of hitting the target by $\mathit{B}=\frac{1}{2}$ $\Rightarrow \mathit{P}\left({\mathit{B}}^{\prime }\right)=1-\mathit{P}\left(\mathit{B}\right)=1-\frac{1}{2}=\frac{1}{2}$
$\mathit{P}\left(\mathit{C}\right)$ : Probability of hitting the target by $\mathit{C}=\frac{1}{3}$ $\Rightarrow \mathit{P}\left({\mathit{C}}^{\prime }\right)=1-\mathit{P}\left(\mathit{C}\right)=1-\frac{1}{3}=\frac{2}{3}$
$\mathit{P}\left(\mathit{E}\right)$ : Probability that A will be hitting
$\Rightarrow \mathit{P}\left(\mathit{E}\right)=1-\mathit{P}\left({\mathit{B}}^{\prime }\right)\mathit{P}\left({\mathit{C}}^{\prime }\right)$
$\Rightarrow \mathit{P}\left(\mathit{E}\right)=1-\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)$
$\Rightarrow \mathit{P}\left(\mathit{E}\right)=\frac{2}{3}$
Probability that $\mathit{A}$ is hit then the probability that $\mathit{B}$ hits the target and $\mathit{C}$ does not $=\mathit{P}\left(\mathit{B}\cap {\mathit{C}}^{\prime }\mid \mathit{E}\right)$ $=\frac{\mathit{P}\left(\mathit{B}\right)\mathit{P}\left({\mathit{C}}^{\prime }\right)}{\mathit{P}\left(\mathit{E}\right)}$

$=\frac{\frac{1}{2}\times \frac{2}{3}}{\frac{2}{3}}$
$=\frac{1}{2}$
Hence, $\mathit{A}$ is targeting $\mathit{B},\mathit{B}$ and $\mathit{C}$ are targeting A. Probability of hitting the target by $\mathit{A},\mathit{B}$ and $\mathit{C}$ are $\frac{2}{3},\frac{1}{2}$ and $\frac{1}{3}$ respectively. If $\mathit{A}$ is hit then the probability that $\mathit{B}$ hits the target and $\mathit{C}$ does not, is $\frac{1}{2}$