Probability that a candidate is placed in the class

Question

Examination candidates are graded into four classes known conventionally as I, II–1, II–2 and III, with probabilities 1/8, 2/8, 3/8 and 2/8 respectively. A candidate who misreads the rubric, — a common event with probability 2/3 —, generally does worse, his or her probabilities being 1/10, 2/10, 4/10 and 3/10.

What is the probability:

(i) that a candidate who reads the rubric correctly is placed in the class II–1?

(ii) that a candidate who is placed in the class II–1 has read the rubric correctly?

Answer 1

This is how I interpret the question.

If there are N candidates, 2N/3 will misread the rubric and N/3 will read it correctly.

Of those who misread it:

(1/10)(2N/3)=N/15 will be graded class I;

(2/10)(2N/3)=2N/15 class II-1;

(4/10)(2N/3)=4N/15 class II-2;

(3/10)(2N/3)=3N/15 class III.

Of those who read it correctly:

(1/8)(N/3)=N/24 will be graded class I;

(2/8)(N/3)=N/12 class II-1;

(3/8)(N/3)=N/8 class II-2;

(2/8)(N/3)=N/12 class III.

(i) Probability=1/12

(ii) Those placed in II-1 consist of 2N/15 of those who misread the rubric and N/12 of those who read it correctly, totalling 2N/15+N/12=(8+5)N/60=13N/60. So the probability=(N/12)/(13N/60)=(5N/60)/(13N/60)=5/13.