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An urn contains bblack balls and rred balls. One of the
Chapter 3, Problem 6STE(choose chapter or problem)
Problem 6STE
An urn contains bblack balls and rred balls. One of the balls is drawn at random, but when it is put back in the urn, cadditional balls of the same color are put in with it. Now, suppose that we draw another ball. Show that the probability that the first ball was black, given that the second ball drawn was red, is b/(b + r + c).
Questions & Answers
QUESTION:
Problem 6STE
An urn contains bblack balls and rred balls. One of the balls is drawn at random, but when it is put back in the urn, cadditional balls of the same color are put in with it. Now, suppose that we draw another ball. Show that the probability that the first ball was black, given that the second ball drawn was red, is b/(b + r + c).
ANSWER:
Step 1 of 2
It is given that there are b black and r red balls in an urn.
Also, it is given that one ball is drawn at random and when it put back in the urn, the conditional balls of the same color are put with it.
Using this we have to show that the probability that the first drawn ball was black given that the second ball drawn was red is .