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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 by Q
Solution
Let E1, E2, E3 be the events that P, Q, R become principal.
E1, E2, E3 are mutually exclusive and exhaustive
It is given that, P(E1) = `2/5`, P(E2) = `2/5`, P(E3) = `1/5`.
Let T ≡ the event that IT is introduced
`"P"("T"/"E"_1)` = Probability that IT is introduced if P becomes Principal
= `1/2` ...(Given)
Also, it is given that
`"P"("T"/"E"_2) = 1/3, "P"("T"/"E"_3) = 1/4`
The required probability = `"P"("E"_2/"T")`
= `("P"("E"_ 2 ∩"T"))/("P"("T"))`
= `("P"("E"_2)*"P"("T"/"E"_2))/("P"("T"))`
= `((2/5)*(1/3))/((23/60))`
= `8/23`
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(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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