Let ‘E’ denotes the event that student passes in English examination.
And ‘H’ be the event that student passes in Hindi exam.
It is given that,
P (E) = 0.75
P (passing both) = P (E ∩ H) = 0.5
P (passing neither) = P (E′ ∩ H′) = 0.1
P (H) =?
As, we know that P (A′ ∩ B′) = P (A ∪ B) ′ {using De Morgan’s law}
P (E′ ∩ H′) = P (E ∪ H)′
0.1 = 1 – P (E ∪ H)
P (E ∪ H) = 1 – 0.1
= 0.9
By using the 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)
0.9 = 0.75 + P (H) – 0.5
1.4 – 0.75 = P (H)
∴ P (H) = 0.65
