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प्रश्न
A box contains 2 blue and 3 pink balls and another box contains 4 blue and 5 pink balls. One ball is drawn at random from one of the two boxes and it is found to be pink. Find the probability that it was drawn from second box
उत्तर
Let E1 ≡ the event that first box is chosen
E2 ≡ the event that second box is chosen
E1, E2 are mutually exclusive and exhaustive.
Also, P(E1) = P(E2) = `1/2`
Let A ≡ the event that pink ball is drawn
The first box contains 2 blue and 3 pink balls
i.e., altogether 5 balls.
∴ `"P"("A"/"E"_1)` = Probability that pink marble is drawn given that first box is chosen
= `3/5`
Similarly, `"P"("A"/"E"_2) = 5/9`
By Boye's Theorem, the required probability
= `"P"("E"_2/"A")`
= `("P"("E"_2)*"P"("A"/"E"_2))/("P"("E"_1)*"P"("A"/"E"_1) + "P"("E"_2)*"P"("A"/"E"_2))`
= `((1/2)*(5/9))/((1/2)*(3/5) + (1/2)*(5/9))`
= `((5/9))/((3/5) + (5/9)`
= `((5/9))/((52/45))`
= `25/52`
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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) = 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`
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`= square`
`"P"("C"//"L") = ("P"("L" ∩ "C"))/("P"("L"))`
= `("P"("C") * "P"("L"//"C"))/("P"("L"))`
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| CASE-BASED/DATA-BASED |
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