Question

If a students selected at random is found to have passed the examination, then the probability that he was the only students who has passed the examination is

• A. $1/3025$
• B. $1/605$
• C. $1/275$
• D. $1/121$$

Answer 1

The correct option is A $1/3025$

Let P(i) be the probability that exactly i students are passing an examination.Now given that

$P\left(A_i\right)=\lambda i^2$ (where $\lambda$ is constant)

$\Rightarrow {\sum }_{i=1}^{10}P\left(A_i\right)={\sum }_{i=1}^{10}\lambda \frac{10\times 11\times 21}{6}=\lambda 385=1\Rightarrow \lambda \Rightarrow 1/385$

Let A represent the event that selected students have passed the examination.

$\therefore P\left(A\right)={\sum }_{i=1}^{10}P\left(A/A_i\right)P\left(A_i\right)$

$={\sum }_{i=1}^{10}\frac{i}{10}\frac{i^1}{385}$

$=\frac{1}{3850}{\sum }_{i=1}^{10}{i}^3$

$=\frac{{10}^2{11}^2}{4\times 3850}=\frac{11}{14}$

Now,

$P\left(A_1/A\right)=\frac{P\left(A/A_1\right)P\left(A_1\right)}{P\left(A\right)}$

$=\frac{\frac{1}{385}\frac{1}{10}}{\frac{11}{14}}$

$=\frac{1}{11\times 55}\frac{1}{5}$

$=\frac{1}{3025}$