Question

There are n students in a class and probability that exactly \[\lambda \] out of n pass the examination is directly proportional to \[{{\lambda }^{2}}\] \[\left( 0\le \lambda \le n \right)\]. Find out the probability that the student selected at random has passed the examination.
\[\begin{align}
  & (A)\text{ }P(A)=\dfrac{3(n+1)}{(2n+1)} \\
 & (B)\text{ }P(A)=\dfrac{(n+1)}{2(2n+1)} \\
 & (C)\text{ }P(A)=\dfrac{3(n+1)}{2(4n+1)} \\
 & (D)\text{ }P(A)=\dfrac{3(n+1)}{2(2n+1)} \\
\end{align}\]

Answer 1

Hint: Let us assume the probability of \[\lambda \] students passed among n students is represented by \[P({{E}_{\lambda }})\] \[\left( 0\le \lambda \le n \right)\]. From the question, it is clear that \[P({{E}_{\lambda }})\] is proportional to \[{{\lambda }^{2}}\]. We know that the sum of probability of occurrence of all the events is equal to 1. So, the sum of all probability of occurrence of all events \[{{E}_{\lambda }}\] in the range \[\left( 0\le \lambda \le n \right)\] is 1. From this we will get a relation between n and k. Now by using the conditional probability concept, we can find the probability that the student at random has passed the examination.
\[P(A)=P({{E}_{1}})P\left( \dfrac{A}{{{E}_{1}}} \right)+P({{E}_{2}})P\left( \dfrac{A}{{{E}_{2}}} \right)+........+P({{E}_{\lambda }})P\left( \dfrac{A}{{{E}_{\lambda }}} \right)\] where \[{{E}_{1}},{{E}_{2}},{{E}_{3}},.....,{{E}_{n}}\] are n events. Here\[P\left( \dfrac{A}{{{E}_{i}}} \right)\] represents the probability of number of students passes the examination, \[P\left( A \right)\]represents the probability of number of students passed the examination. From the relation between n and k, we can obtain the final value of probability that the student selected at random has passed the examination.

Complete step-by-step answer:
Before solving the question, we needed to understand the definition of probability. The probability is the ratio of number of favourable outcomes to total number of outcomes.
In the question, it is given that there are n students in a class in which only \[\lambda \] students out of n students pass the examination.
Probability of \[\lambda \] students among n students passing the examination is directly proportional to \[{{\lambda }^{2}}\].
If \[\lambda \] students passed the examination, assume the event of occurred is represented by \[{{E}_{\lambda }}\]where \[\left( 0\le \lambda \le n \right)\] and also assume the probability of \[\lambda \] students passed among n students is represented by \[P({{E}_{\lambda }})\].
Hence, we can write
\[\Rightarrow P({{E}_{\lambda }})\propto {{\lambda }^{2}}\]
\[\Rightarrow P({{E}_{\lambda }})=k{{\lambda }^{2}}......(1)\] where k is proportionality constant.
We know that the sum of probability of occurrence of all the events is equal to 1.
So, the sum of all probability of occurrence of all events \[{{E}_{\lambda }}\] in the range \[\left( 0\le \lambda \le n \right)\] is 1.
\[\begin{align}
  & P({{E}_{1}})+P({{E}_{2}})+.......+P({{E}_{n}})=1 \\
 & \Rightarrow k{{(1)}^{2}}+k{{(2)}^{2}}+....+k{{(n)}^{2}}=1 \\
 & \Rightarrow k\left( \dfrac{n(n+1)(2n+1)}{6} \right)=1 \\
\end{align}\]
\[\Rightarrow k=\dfrac{6}{n(n+1)(2n+1)}......(2)\]
Assume the probability that a student passes the examination is \[P(A)\].
We know that from the concept of conditional probability,
\[P(A)=P({{E}_{1}})P\left( \dfrac{A}{{{E}_{1}}} \right)+P({{E}_{2}})P\left( \dfrac{A}{{{E}_{2}}} \right)+........+P({{E}_{n}})P\left( \dfrac{A}{{{E}_{n}}} \right)=\sum\limits_{\lambda =1}^{n}{P({{E}_{\lambda }}})P\left( \dfrac{A}{{{E}_{\lambda }}} \right).....(3)\]
where \[P({{E}_{\lambda }})=k{{\lambda }^{2}}\] and \[P\left( \dfrac{A}{{{E}_{\lambda }}} \right)\] represents the probability of number of students passes the examination.
Hence, \[P\left( \dfrac{A}{{{E}_{\lambda }}} \right)=\dfrac{\lambda }{n}....(4)\]
Substitute equation (1) and equation (4) in equation (3).
\[\Rightarrow P(A)=\left( k{{(1)}^{2}} \right)\left( \dfrac{1}{n} \right)+\left( k{{(2)}^{2}} \right)\left( \dfrac{2}{n} \right)+....\left( k{{(n)}^{2}} \right)\left( \dfrac{n}{n} \right)\]
\[\Rightarrow P(A)=\dfrac{k}{n}{{(1)}^{3}}+\dfrac{k}{n}{{(2)}^{3}}+.......+\dfrac{k}{n}{{(n)}^{3}}\]
\[\Rightarrow P(A)=\dfrac{k}{n}\left( {{1}^{3}}+{{2}^{3}}+.....{{n}^{3}} \right)\]
We know that sum of cubes of natural numbers = \[\left( {{1}^{3}}+{{2}^{3}}+.....{{n}^{3}} \right)={{\left( \dfrac{n(n+1)}{2} \right)}^{2}}\]
\[\Rightarrow P(A)=\dfrac{k}{n}{{\left( \dfrac{n(n+1)}{2} \right)}^{2}}......(5)\]
Substitute equation (2) in equation (5).
\[\begin{align}
  & \Rightarrow P(A)=\dfrac{1}{n}\left( \dfrac{6}{n(n+1)(2n+1)} \right)\left( \dfrac{n(n+1)}{2} \right)\left( \dfrac{n(n+1)}{2} \right) \\
 & \Rightarrow P(A)=\dfrac{6(n)(n+1)(n)(n+1)}{(n)(n)(n+1)(2n+1)(2)(2)} \\
 & \Rightarrow P(A)=\dfrac{6{{(n)}^{2}}{{(n+1)}^{2}}}{4{{(n)}^{2}}(n+1)(2n+1)} \\
 & \Rightarrow P(A)=\dfrac{6(n+1)}{4(2n+1)} \\
 & \Rightarrow P(A)=\dfrac{3(n+1)}{2(2n+1)}....(6) \\
\end{align}\]
Therefore, the probability that the student selected at random has passed the examination is \[P(A)=\dfrac{3(n+1)}{2(2n+1)}\].
Hence, option (4) is correct.

Note: One should be careful while understanding the difference between the terms\[P\left( \dfrac{A}{{{E}_{\lambda }}} \right)\] and \[P\left( \dfrac{{{E}_{\lambda }}}{A} \right)\] . \[P\left( \dfrac{A}{{{E}_{\lambda }}} \right)\]represents the probability of occurrence of event A after the occurrence of event \[{{E}_{\lambda }}\] and \[P\left( \dfrac{{{E}_{\lambda }}}{A} \right)\] represents the probability of occurrence of event \[{{E}_{\lambda }}\] after the occurrence of event A. If a small mistake is done while understanding the difference, the solution will become incorrect.