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) = \(\frac{n(E)}{n(S)}\) = \(\frac{60}{100}\) = \(\frac{3}{5}\) 

Similarly, P(H) =  \(\frac{n(H)}{n(S)}\) = \(\frac{50}{100}\) = \(\frac{1}{2}\)  

And, P(E∩H) = \(\frac{P(E∩H)}{n(S)}\) = \(\frac{30}{100}\) = \(\frac{3}{10}\) 

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) = \(\frac{3}{5}+\frac{1}{2}-\frac{3}{10}\) = \(\frac{11}{10}-\frac{3}{10}\) 

= \(\frac{8}{10} = \frac{4}{5} \) 

∴ P(E∪H) = \(\frac{4}{5}\)