Question

If a selected student has been found to pass the examination then find out the probability that he is the only student to have passed the examination.

• A. $P\left(E_1/A\right)=\frac{4}{{\left[n\right(2n+1\left)\right]}^2}$
• B. $P\left(E_1/A\right)=\frac{2}{{\left[n\right(2n+1\left)\right]}^2}$
• C. $P\left(E_1/A\right)=\frac{4}{{\left[n\right(n+1\left)\right]}^2}$
• D. $P\left(E_1/A\right)=\frac{2}{{\left[n\right(n+1\left)\right]}^2}$

Answer 1

The correct option is C $P\left(E_1/A\right)=\frac{4}{{\left[n\right(n+1\left)\right]}^2}$

Let $E_{\lambda }$ be the event that exactly $\lambda$ out of n pass the examination and
$A$ is the event that a student pass the examination
$P\left(E_{\lambda }\right)\propto {\lambda }^2\Rightarrow P\left(E_{\lambda }\right)={k\lambda }^2$ (k os proportionality constant)
$\therefore E_0,E_1,E_2,...E_n$ are mutually exclusive and exhaustive events.
$\Rightarrow P\left(E_0\right)+P\left(E_1\right)+P\left(E_2\right)+...P\left(E_n\right)=1\Rightarrow 0+k{\left(1\right)}^2+k{\left(2\right)}^2+...+k{\left(n\right)}^2=1\Rightarrow k(1^2+2^2+3^2+...n^2)=1$
$\Rightarrow \frac{kn(n+1)(2n+1)}{6}=1\Rightarrow k=\frac{6}{n(n+1)(2n+1)}$
$P\left(A\right)={\sum }_{r=1}^{n}P\left(E_{\lambda }\right)P\left(\frac{A}{E_{\lambda }}\right)={\sum }_{r=1}^{n}k{\lambda }^2\times \frac{\lambda }{n}=\frac{k}{n}{\sum }_{r=1}^n{\lambda }^3$
$=\frac{k}{n}{\left[\frac{n(n+1)}{2}\right]}^2=\frac{6}{n^2(n+1)(2n+1)}.\frac{n^2{(n+1)}^2}{4}=\frac{3(n+1)}{2(2n+1)}$
$P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right)P\left(\frac{A}{E_1}\right)}{P\left(A\right)}=\frac{\frac{6}{n(n+1)(2n+1)}\times \frac{1}{n}}{\frac{3(n+1)}{2(2n+1)}}=\frac{4}{{\left[n(n+1)\right]}^2}$