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Hundred students appeared for two examinations. 60 passed the first, 50 passed the second, and 30 passed in both. Find the probability that student selected at random failed in both the examinations.

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#### Solution

Out of a hundred students, 1 student can be selected in ^{100}C_{1} = 100 ways.

∴ n(S) = 100

Let A be the event that student passed in the first examination.

Let B be the event that student passed in the second examination.

∴ n(A) = 60, n(B) = 50 and n(A ∩ B) = 30

∴ P(A) = `("n"("A"))/("n"("S")) = 60/100 = 6/10`

∴ P(B) = `("n"("B"))/("n"("S")) = 50/100 = 5/10`

∴ P(A ∩ B) = `("n"("A" ∩ "B"))/("n"("S")) = 30/100 = 3/10`

P(student failed in both examinations)

= P(A' ∩ B')

= P(A ∪ B)' .......[De Morgan's law]

= 1 – P(A ∪ B)

= 1 − P(A) + P(B) − P(A ∩ B)

= `1-6/10+5/10-3/10`

= `1-8/10`

= `1 - 4/5`

= `1/5`

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