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.