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

Step 1: Understand the problem
- 20% of students have above 90% attendance.
- 80% of students are irregular.
- 80% of students with above 90% attendance get an 'A' grade.
- 20% of irregular students get an 'A' grade.
A student chosen at random has an 'A' grade. Find the probability that this student is irregular.

Step 2: Define events
Let:
A = event that the student has an 'A' grade.
H = event that the student has above 90% attendance.
I = event that the student is irregular.

Step 3: Given probabilities
P(H) = 0.20 (students with high attendance)
P(I) = 0.80 (irregular students)
P(A|H) = 0.80 (probability of 'A' given high attendance)
P(A|I) = 0.20 (probability of 'A' given irregular)

Step 4: Calculate total probability of 'A'
Using 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

Step 5: Use Bayes' theorem to find P(I|A)
\[ P(I|A) = \frac{P(A|I) \times P(I)}{P(A)} = \frac{0.20 \times 0.80}{0.32} = \frac{0.16}{0.32} = 0.5 \]

Step 6: Conclusion
The probability that the student is irregular given that the student got an 'A' grade is 0.5 or 1/2.

Final Answer: 1/2