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Conditional Probability

Three bags co...

Question

Three bags contain $2$ silver, $5$ copper coins, and $3$ silver, $4$ copper coins and $5$ silver, $2$ copper coins respectively. A bag is chosen at random and a coin is drawn from it which happens to be silver. What is the probability that it has come from third bag?

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Solution

$U s i n g B a y e^{'} s R u l e, S e l e c t i o n o f e a c h b a g i s e q u a l l y l i k e l y w i t h p r o b a b i l i t y 1 / 3 P r o b a b i l i t y t h a t i t s c o m i n g f r o m t h e t h i r d b a g g i v e n i t s s i l v e r = \frac{P r o b a b i l i t y t h a t i t s c o m i n g f r o m t h e t h i r d b a g g i v e n i t s s i l v e r}{P r o b a b i l i t y t h a t t h e c o i n i s s i l v e r f r o m a n y b a g} = \frac{(1 / 3) \times (5 / 7)}{(1 / 3) \times (2 / 7) + (1 / 3) \times (3 / 7) + (1 / 3) \times (5 / 7)} = \frac{5}{10} = 1 / 2$

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