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Question

Bag -I contains 3 red and 4 black balls while another Bag -II contains 5 red and 6 black b alls . One ball is drawn at random from one of the bags and it is found to be red . Find the probability that it was drawn from Bag -II.


Solution

Let $${E_1}\,,{E_2}$$ be the event of choosing bag 1 and bag 2 respectively.
$$E$$ be the event that the drawn ball is red.
Then,

$$\begin{array}{l} P\left( { { E_{ 1 } } } \right) =P\left( { { E_{ 2 } } } \right) =\frac { 1 }{ 2 }  \\ P\left( { E/{ E_{ 1 } } } \right) =\frac { 3 }{ 7 }  \\ P\left( { E/{ E_{ 2 } } } \right) =\frac { 5 }{ { 11 } }  \\ Apply\, the\, Bayes\, theorem \\ P\left( { { E_{ 2 } }/{ E_{ 1 } } } \right) =\frac { { \frac { 1 }{ 2 } \times \frac { 5 }{ { 11 } }  } }{ { \frac { 1 }{ 2 } \times \frac { 3 }{ 7 } +\frac { 1 }{ 2 } \times \frac { 5 }{ { 11 } }  } }  \\ =\frac { { 35 } }{ { 68 } }  \end{array}$$  

Mathematics

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