Question

It is known that 20% of the students in a school have above 90% attendance and 80% of the students are irregular. Past year results show that 80% of students who have above 90% attendance and 20% of irregular students get “A” grade in their annual examination. At the end of a year, a student is chosen at random from the school and is found to have an “A” grade. What is the probability that the student is irregular?

Answer 1

Problem Statement: A randomly selected student has received an 'A' grade. Given that 20% of students have above 90% attendance and 80% are irregular, and that 80% of students with above 90% attendance and 20% of irregular students receive an 'A' grade, determine the probability that this student is irregular.

Event Definitions:

• A = Student receives an 'A' grade.
• H = Student has above 90% attendance.
• I = Student is irregular.

Given Probabilities:

• P(H) = 0.20
• P(I) = 0.80
• P(A|H) = 0.80
• P(A|I) = 0.20

Calculation of Total Probability of 'A':

Using the law of total probability:

P(A) = P(A|H)P(H) + P(A|I)P(I)
= (0.80)(0.20) + (0.20)(0.80)
= 0.16 + 0.16 = 0.32

Application of Bayes' Theorem:

To find the probability that a student is irregular given they received an 'A' grade (P(I|A)):

\[P(I|A) = \frac{P(A|I) \times P(I)}{P(A)} = \frac{0.20 \times 0.80}{0.32} = \frac{0.16}{0.32} = 0.5\]

Conclusion:

The probability that a student who received an 'A' grade is irregular is 0.5.

Final Answer: 1/2