Let E denotes the event that student passed in first examination.

And H be the event that student passed in second exam.

S is the sample space containing the students who appeared for the exam.

Given,

n(S) = 100

n(E) = 60

n(H) = 50

also no of students who passed both exam = n(E ∩ H) = 30

∴ P(E) =

Similarly, P(H) =

And, P(E ∩ H) =

We need to find the probability of event such that a student selected at random has passed at least one examination.

This can be given as – P(E or H) = P(E ∪ H)

Note: By definition of P(A or B) under axiomatic approach(also called addition theorem) we know that:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴ P(E ∪ H) = P(E) + P(H) – P(E ∩ H)

⇒ P(E ∪ H) =

∴ P(E ∪ H) = 4/5