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Question
There are two identical urns containing respectively 6 black and 4 red balls, 2 black and 2 red balls. An urn is chosen at random and a ball is drawn from it. if the ball is black, what is the probability that it is from the first urn?
Solution
| Black balls | Red balls | Total | |
| Urn I | 6 | 4 | 10 |
| Urn I | 2 | 2 | 4 |
| Total | 8 | 6 | 14 |
The conditional Probability of A1 given B is P(A1/B)
By Bayes’ theorem
P(A1/B) = `("P"("A"_1) * "P"("B"/"A"_1))/("P"("A"_1) * "P"("B"/"A"_1) + "P"("A"_2) * "P"("B"/"A"_2))`
= `(1/2 xx 3/5)/(1/2 xx 3/5 + 1/2 xx 1/2)`
= `(3/10)/(3/10 + 1/4)`
= `(3/10)/((6 + 5)/20)`
P(A1/B) = `3/10 xx 20/11`
= `6/11`
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