Of the students in a school, it is known that 30% have 100% attendance and 70% students are irregular. Previous year results report that 70% of all students who have 100% attendance attain A grade and 10% irregular students attain A grade in their annual examination. At the end of the year, one student is chosen at random from the school and he was found to have an A grade. What is the probability that the student has 100% attendance? Is regularity required only in school? Justify your answer.

Let E₁ : Student has 100% attendance
E₂ : Student is irregular
A: Student attains A grade
$\frac{\mathrm{A}}{{\mathrm{E}}_1}$ : Student attains A grade given that he has $100 \%$ attendance
$\frac{\mathrm{A}}{{\mathrm{E}}_2}$: Student attains A grade given that he is irregular
$\frac{{\mathrm{E}}_1}{\mathrm{A}}:$ Student has 100% attendance given that he attains A grade

Using Bayes' theorem
$\begin{array}{l}\mathrm{P}\left(\frac{{\mathrm{E}}_1}{\mathrm{A}}\right)=\frac{\mathrm{P}\left({\mathrm{E}}_1\right)\cdot \mathrm{P}\left(\frac{\mathrm{A}}{{\mathrm{E}}_1}\right)}{\mathrm{P}\left({\mathrm{E}}_1\right)\cdot \mathrm{P}\left(\frac{\mathrm{A}}{{\mathrm{E}}_1}\right)+\mathrm{P}\left({\mathrm{E}}_2\right)\cdot \mathrm{P}\left(\frac{\mathrm{A}}{{\mathrm{E}}_2}\right)}\\ =\frac{\frac{30}{100}\times \frac{70}{100}}{\frac{30}{100}\times \frac{70}{100}+\frac{70}{100}\times \frac{10}{100}}\\ =\frac{30\times 70}{30\times 70+70\times 10}\\ =\frac{30\times 70}{70(30+10)}\\ =\frac{30}{30+10}\\ =\frac{30}{40}\\ =\frac{3}{4}\end{array}$

Regularity is the value which is required at every stage of our life. In our childhood, during school education we can inculcate this value in our personality. Regularity increases our capabilities and makes us able to put the best of our potential. We are able to achieve certain targets due to regular efforts.