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?

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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