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Question
Bag I contains 3 Red and 4 Black balls while another Bag II contains 5 Red and 6 Black balls. 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 I.
Solution
| Red | Black | Total | |
| Urn I | 3 | 4 | 7 |
| Urn II | 5 | 6 | 11 |
Let E1 be the event of choosing the first bag, E2 be the event of choosing the second bag.
Let A be the event of drawing a red ball.
Then P(E1) = `1/2`, P(E2) = `1/2`
Also P`("A"/"E"_1)` = P(Drawing a red ball from the bag I) = `3/7`
P`("A"/"E"_2)` = P(Drawing a red ball from bag II) = `5/11`
The probability of drawing a ball from bag I, being given that it is red, is P`("E"_1/"A")`
`"P"("E"_1/"A") = ("P"("E"_1) "P"(("A")/("E"_1)))/("P"("E"_1) "P"(("A")/("E"_1)) + "P"("E"_2) "P"(("A")/"E"_2)`
= `(1/2 xx 3/7)/(1/2 xx 3/7 + 1/2 xx 5/11)`
= `(3/7)/(3/7 + 5/11)`
= `(3/7)/(((33 + 35)/(7 xx 11)))`
= `3/(7 xx 68) xx 7 xx 11`
= `33/68`
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