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Question
There are three categories of students in a class of 60 students:
A : Very hardworking ; B : Regular but not so hardworking; C : Careless and irregular 10 students are in category A, 30 in category B and the rest in category C. It is found that the probability of students of category A, unable to get good marks in the final year examination is 0.002, of category B it is 0.02 and of category C, this probability is 0.20. A student selected at random was found to be one who could not get good marks in the examination. Find the probability that this student is category C.
Solution
Let E denote the event that the student could not get good marks in the examination.
Also, A : the event that student is very hardworking
B : the event that student is regular but not so hardworking
C : the event that student is careless and irregular
∴ P(A) = \[\frac{10}{60}\] , P(B) = \[\frac{30}{60}\] and P(C) = \[\frac{20}{60}\]
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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`
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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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