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Question
Solve the following:
The chances of P, Q and R, getting selected as principal of a college are `2/5, 2/5, 1/5` respectively. Their chances of introducing IT in the college are `1/2, 1/3, 1/4` respectively. Find the probability that IT is introduced in the college after one of them is selected as a principal
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Solution
Let event P: P become principal.
event Q: Q become principal.
event R: R become principal.
event E: Subject IT is introduced.
Given, `"P"("P") = 2/5, "P"("Q") = 2/5, "P"("R") = 1/5`
and `"P"("E"/"P") = 1/2, "P"("E"/"Q") = 1/3`,
`"P"("E"/"R") = 1/4`
Required probability = P(E)
= `"P"("P") * "P"("E"/"P") + "P"("Q") * "P"("E"/"Q") + "P"("R") * "P"("E"/"R")`
= `2/5 xx 1/2 + 2/5 xx 1/3 + 1/5 xx 1/4`
= `1/5 + 2/15 + 1/20`
= `(12 + 8 + 3)/60`
= `23/60`
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Solution: Let A, C and T be the events that Mr. X goes to office by Auto, Car and Train respectively. Let L be event that he is late.
Given that P(A) = `square`, P(C) = `square`
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P(L/A) = `1/2`, P(L/C) = `square` P(L/T) = `1/4`
P(L) = P(A ∩ L) + P(C ∩ L) + P(T ∩ L)
`="P"("A")*"P"("L"//"A") + "P"("C")*"P"("L"//"C") + "P"("T")*"P"("L"//"T")`
`= square * square + square * square + square * square`
`= square + square + square`
`= square`
`"P"("C"//"L") = ("P"("L" ∩ "C"))/("P"("L"))`
= `("P"("C") * "P"("L"//"C"))/("P"("L"))`
`= (square * square)/square`
`= square`
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