In an entrance test that is graded on the basis of two examinations, t

1.	In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of a the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both?
Answer» 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

In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of a the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both?

Answer»

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