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Question
(Activity):
Mr. X goes to office by Auto, Car, and train. The probabilities him travelling by these modes are `2/7, 3/7, 2/7` respectively. The chances of him being late to the office are `1/2, 1/4, 1/4` respectively by Auto, Car, and train. On one particular day, he was late to the office. Find the probability that he travelled by car.
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`
P(T) = `square`
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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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) = `underline(2/7)`, P(C) = `underline(3/7)`, P(T) = `underline(2/7)`
`"P"("L"//"A") = 1/2, "P"("L"//"C") = underline(1/4), "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")`
= `underline(2/7) xx underline(1/2) + underline(3/7) xx underline(1/4) + underline(2/7) xx underline(1/4)`
= `underline(1/7) + underline(3/28) + underline(2/28)`
`= underline(9/28)`
`"P"("C"//"L") = ("P"("L" ∩ "C"))/("P"("L"))`
= `("P"("C") * "P"("L"//"C"))/("P"("L"))`
= `(underline(3/7) xx underline(1/4))/(underline(9/28)`
= `underline(1/3)`
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