Question

There are $n$ students in a class and probability that exactly $\lambda$ out of $n$ pass the examination is directly proportional to ${\lambda }^2(0\le \lambda \le n)$. Find out the probability that a student selected at random has passed the examination.

• A. $P\left(A\right)=\frac{3(n+1)}{(2n+1)}$
• B. $P\left(A\right)=\frac{(n+1)}{2(2n+1)}$
• C. $P\left(A\right)=\frac{3(n+1)}{2(4n+1)}$
• D. $P\left(A\right)=\frac{3(n+1)}{2(2n+1)}$

Answer 1

The correct option is D $P\left(A\right)=\frac{3(n+1)}{2(2n+1)}$

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 is proportionality constant)
$\therefore E_0,E_1,E_2,...E_n$ are mutually exclusive and exhaustive events.
$\Rightarrow P\left(E_0\right)+P\left(E_1\right)+P\left(E_2\right)+...P\left(E_n\right)=1\Rightarrow 0+k{\left(1\right)}^2+k{\left(2\right)}^2+...+k{\left(n\right)}^2=1\Rightarrow k(1^2+2^2+3^2+...n^2)=1$
$\Rightarrow \frac{kn(n+1)(2n+1)}{6}=1\Rightarrow k=\frac{6}{n(n+1)(2n+1)}$
$P\left(A\right)={\sum }_{\lambda =1}^{n}P\left(E_{\lambda }\right)P\left(\frac{A}{E_{\lambda }}\right)={\sum }_{\lambda =1}^{n}k{\lambda }^2\times \frac{\lambda }{n}=\frac{k}{n}{\sum }_{\lambda =1}^n{\lambda }^3$
$=\frac{k}{n}{\left[\frac{n(n+1)}{2}\right]}^2=\frac{6}{n^2(n+1)(2n+1)}.\frac{n^2{(n+1)}^2}{4}=\frac{3(n+1)}{2(2n+1)}$