Let A: student passes first examination B: student passes second examination 

Given: P(A) = 0.8, P(B) = 0 .7 and 

P(A∪B) = 0.95 

To find: P(A∩B) 

∵ P(A∩B) = P(A) + P(B) – P(A∪B) = 0.55

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

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

Given, P(E) = 0.8 and P(H) = 0.7 

Also probability of passing atleast one exam i.e P(E or H) = 0.95 

Or, P(E∪H) = 0.95 

We have to find the probability of the event in which students pass both the examinations i.e. 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) + P(H) – P(E∪H) 

⇒ P(E∩H) = 0.7 + 0.8 – 0.95 

= 1.5 – 0.95 = 0.55 

∴ Probability of passing both the exams = P(E∩H) = 0.55