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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 - Mathematics and Statistics

Sum

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

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Solution

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`

Concept: Baye'S Theorem
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