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प्रश्न
Of the students in a school, it is known that 30% have 100% attendance and 70% students are irregular. Previous year results report that 70% of all students who have 100% attendance attain A grade and 10% irregular students attain A grade in their annual examination. At the end of the year, one student is chos~n at random from the school and he was found ·to have an A grade. What is the probability that the student has 100% attendance? Is regularity required only in school? Justify your answer
उत्तर
Let `E_1` is event of students which have 100% attendance and `E_2` is event of students which are Irregular
then `P(E_1) = 0.3`
`P(E_2) = 0.7`
Let A : Event of students which attendance A grade
then P(A/ E1) = 0.7 and P(A/ E2) = 0.1
So By Bays theorem
P(Probability that student has 100% Attendance)
As per answer, the probability of regular students is more than 50%. So the regularty is required.
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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`
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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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