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Question
Given three identical Boxes A, B and C, Box A contains 2 gold and 1 silver coin, Box B contains 1 gold and 2 silver coins and Box C contains 3 silver coins. A person choose a Box at random and takes out a coin. If the coin drawn is of silver, find the probability that it has been drawn from the Box which has the remaining two coins also of silver.
Solution
Using Baye’s theorem
`therefore P(C/S) = P(C)xx(P(S/C))/(P(A)xxP(B)xxP(S/B)+P(C)xxP(S/C))`
= `(1/3 xx 3/3)/(1/3xx1/3+1/3xx2/3 +1/3xx3/3)`
= `1/(1/3+2/3+3/3`
= `1/(6/3)`
`1/2`
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