In grade 11 of a school 40 students opted for Physics 17 opted for Biology and 20 opted for Chemistry. If the total number of students in grade 11 was 60 all of these students opted for at least one of the three subjects mentioned here and exactly five of these students opted for all these three subjects what is the probability that a randomly selected student of grade 11 of this school would have opted for exactly one of these three subjects?

Correct option is D

Given:
Number of students who opted for Physics (P) = 40
Number of students who opted for Biology (B) = 17
Number of students who opted for Chemistry (C) = 20
Total number of students [n(P ∪ B ∪ C)] = 60
Number of students who opted for all three subjects (P ∩ B ∩ C) = 5
Solution:
Using the principle of inclusion-exclusion,
n(P ∪ B ∪ C)] = n(P) + n(B) + n(C) - n(P ∩ B) - n(P ∩ C) - n(B ∩ C) + n(P ∩ B ∩ C)
60 = 40 + 17 + 20 - n(P ∩ B) - n(P ∩ C) - n(B ∩ C) + 5
n(P ∩ B) + n(P ∩ C) + n(B ∩ C) = 22
n(exactly one subject) = n(P) + n(B) + n(C) - 2 × (n(P ∩ B) )+ n(P ∩ C) + n(B ∩ C)) + 3 × n(P ∩ B ∩ C)
n(exactly one subject) = 40 + 17 + 20 - 2 × 22 + 15
n(exactly one subject) = 48
Probability = n(exactly one subject) / n(total students)
Probability = 48 / 60 = 0.80