Question

In a class of 75 students, 15 are above average, 45 are average and the rest are below average achievers. The probability that an above average achieving student fails is 0.005, that an average achieving student fails is 0.05 and the probability of a below average achieving student failing is 0.15. If a student is known to have passed, what is the probability that he is a below average achiever?

Answer 1

For answering this question we will use the Bayes’ theorem which states that the probability of an event at a particular condition is given by $P\left( \dfrac{{{A}_{i}}}{E} \right)=\dfrac{P\left( {{A}_{i}} \right)\times P\left( \dfrac{E}{{{A}_{i}}} \right)}{\sum\limits_{i=1}^{n}{P\left( {{A}_{i}} \right)\times P\left( \dfrac{E}{{{A}_{i}}} \right)}}$ . We will take the probability of an above average achiever as $P\left( {{A}_{1}} \right)$, the probability of an average achiever as $P\left( {{A}_{2}} \right)$ and the probability of a below average achiever as $P\left( {{A}_{3}} \right)$. We will use the fact that total probability is 1. Then, we will simplify and derive the required answer.

Complete step by step answer:
In the question it is given that in a class of 75 students, 15 are above average, 45 are average and the rest are below average achievers.
So the probability of an above average achiever is $P\left( {{A}_{1}} \right)=\dfrac{15}{75}$ .
So the probability of an average achiever is $P\left( {{A}_{2}} \right)=\dfrac{45}{75}$
So the probability of a below average achiever is $P\left( {{A}_{3}} \right)=\dfrac{15}{75}$.
Now considering from the question we have the probability that an above average achieving student fails is 0.005, that an average achieving student fails is 0.05 and the probability of a below average achieving student failing is 0.15.
From the statement that the total probability is equal to 1.
So we can say that the probability that an above average achieving student passes is $1-0.005=0.995$ .
So we can say that the probability that an average achieving student passes is $1-0.05=0.95$.
So we can say that the probability that a below average achieving student passes is $1-0.15=0.85$ .
Now by using the Bayes’ theorem, this states that the probability of an event at a particular condition is given by $P\left( \dfrac{{{A}_{i}}}{E} \right)=\dfrac{P\left( {{A}_{i}} \right)\times P\left( \dfrac{E}{{{A}_{i}}} \right)}{\sum\limits_{i=1}^{n}{P\left( {{A}_{i}} \right)\times P\left( \dfrac{E}{{{A}_{i}}} \right)}}$ .
Here in this case we assume the event of the qualified student as $E$ and the probability of above achiever as ${{A}_{1}}$and average achiever as ${{A}_{2}}$ and below average achiever as ${{A}_{3}}$ .
If a student is known to have passed, the probability of that he is a below average achiever is mathematically given by $P\left( \dfrac{{{A}_{3}}}{E} \right)=\dfrac{P\left( {{A}_{3}} \right)\times P\left( \dfrac{E}{{{A}_{3}}} \right)}{\sum\limits_{i=1}^{3}{P\left( {{A}_{i}} \right)\times P\left( \dfrac{E}{{{A}_{i}}} \right)}}$ .
By using this we will have $P\left( \dfrac{{{A}_{3}}}{E} \right)=\dfrac{0.85\times \dfrac{15}{75}}{0.85\times \dfrac{15}{75}+0.995\times \dfrac{15}{75}+0.95\times \dfrac{45}{75}}$ .
By simplifying this we will have $P\left( \dfrac{{{A}_{3}}}{E} \right)=\dfrac{0.85\times \dfrac{1}{5}}{0.85\times \dfrac{1}{5}+0.995\times \dfrac{1}{5}+0.95\times \dfrac{3}{5}}$ .
By performing further simplifications we will have $P\left( \dfrac{{{A}_{3}}}{E} \right)=\dfrac{0.85}{0.85+0.995+0.95\times 3}$ .
By performing calculations we will have $P\left( \dfrac{{{A}_{3}}}{E} \right)=\dfrac{0.85}{1.845+2.85}\Rightarrow \dfrac{0.85}{4.675}$ .

Hence the answer is that in a class of 75 students, 15 are above average, 45 are average and the rest are below average achievers. The probability that an above average achieving student fails is 0.005, that an average achieving student fails is 0.05 and the probability of a below average achieving student failing is 0.15, if a student is known to have passed, the probability of that he is a below average achiever is $0.018$ .

Note: While answering this type of questions we should take care about the calculations if we make mistake during the calculations like taking it as $0.95\times 3$ as $0.285$ instead of $2.85$ and write it as $P\left( \dfrac{{{A}_{3}}}{E} \right)=\dfrac{0.85}{1.845+0.285}\Rightarrow \dfrac{0.85}{2.11}$ the answer we get will be wrong that is $0.402$ .