Question

In a class of $10$ students, probability of exactly $i$ students passing examination is directly proportional to $i^2$. If a student selected at random is found to have passed the examination, then the probability that he was the only student who has passed the examination is

• A. $\frac{1}{3025}$
• B. $\frac{1}{605}$
• C. $\frac{1}{275}$
• D. $\frac{1}{121}$

Answer 1

Answer: (A)

$\frac{1}{3025}$

Let constant of proportionality be $k$.

Let the event that $n$ students pass the exam be $E_n$

Thus, $E_n=k{n}^2$

Thus, $E_0+E_1+E_2...E_{10}=1$

Let $A$ be the event that a student passes the exam.

We need to know $P\left(\frac{E_1}{A}\right)$

Now,

$P\left(A\right)=P\left(E_1\right)P\left(\frac{A}{E_1}\right)+P\left(E_2\right)P\left(\frac{A}{E_2}\right)+...+P\left(E_{10}\right)P\left(\frac{A}{E_{10}}\right)$

Thus,

$P\left(A\right)=\left(k{1}^2\frac{1}{10}\right)+\left(k{2}^2\frac{2}{10}\right)+...+\left(k{9}^2\frac{9}{10}\right)+\left(k{10}^2\frac{10}{10}\right)=\frac{k}{10}(1^3+2^3+...+{10}^3)$

Thus,

$P\left(A\right)=\frac{k}{10}{\left(\frac{10(10+1)}{2}\right)}^2=\frac{3025k}{10}$

Thus,

$P\left(\frac{E_1}{A}\right)=\frac{P\left(E_1\right)P\left(\frac{A}{E_1}\right)}{P\left(A\right)}=\frac{\left(k{1}^2\right)\left(\frac{1}{10}\right)}{\frac{3025k}{10}}=\frac{1}{3025}$