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There are three bags, each containing 100 marbles. Bag 1 has 75 red and 25 blue marbles. Bag 2 has 60 red and 40 blue marbles and Bag 3 has 45 red and 55 blue marbles. One of the bags is chosen - Mathematics and Statistics

Sum

There are three bags, each containing 100 marbles. Bag 1 has 75 red and 25 blue marbles. Bag 2 has 60 red and 40 blue marbles and Bag 3 has 45 red and 55 blue marbles. One of the bags is chosen at random and a marble is picked from the chosen bag. What is the probability that the chosen marble is red?

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Solution

Let B1, B2, B3 be the events that bag 1, bag 2, bag 3 are selected.

Clearly P(B1) = P(B2) = P(B3) =`1/3`.

Let R1, R2, R3 be the events that red marble is drawn from bag 1, bag 2, bag 3 respectively.

If R is the event that red ball is drawn, then

R = (B1 ∩ R1) ∪ (B2 ∩ R2) ∪ (B3 ∩ R3)

The events in the brackets are mutually exclusive

∴ P(R) = P(B1 ∩ R1) + P(B2 ∩ R2) + P(B3 ∩ R3)

= `"P"("B"_1)*"P"("R"_1/"B"_1)+"P"("B"_2)*"P"("R"_2/"B"_2) + "P"("B"_3)* "P"("R"_3/"B"_3)`  ...(1)

`"P"("R"_1/"B"_1)` = Probability that red marble is drawn given that bag 1 is chosen

= `75/100       ...[("Bag 1 has 100 marbles"),("of which 75 are red")]`

Similarly `"P"("R"_2/"B"_2) = 60/100`

`"P"("R"_3/"B"_3) = 45/100`

∴ from (1), P(R) = `1/3* 75/100 + 1/3* 60/100 + 1/3 * 45/100`

=`180/300`

= `3/5`

Concept: Baye'S Theorem
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