An urn contains three red and two white balls. A ball is drawn, and then it and another ball of the same color are placed back in the urn. Finally, a second ball is drawn. a. What is the probability that the second ball drawn is white? b. if the second ball drawn is white, what is the probability that the first ball drawn was red?

A First Course in Probability (10th Edition)
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Chapter1: Combinatorial Analysis
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Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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My question is about finding the answer to b.

Why is drawing a red first not independent of the second draw being a white ball?

 

So, to me, the answer should be 3/5 no matter what ball color the second draw is. Why is that wrong?

I can sort of understand why you find the probability of drawing a red and white by multplying those probabilities together, but why do you divide by the probability of the second ball being white?; Why are we dividing by 2/5?  

An urn contains three red and two white balls. A ball is drawn,
and then it and another ball of the same color are placed back in
the urn.
Finally, a second ball is drawn.
a. What is the probability that the second ball drawn is
white?
b. if the second ball drawn is white, what is the probability
that the first ball drawn was red?
Transcribed Image Text:An urn contains three red and two white balls. A ball is drawn, and then it and another ball of the same color are placed back in the urn. Finally, a second ball is drawn. a. What is the probability that the second ball drawn is white? b. if the second ball drawn is white, what is the probability that the first ball drawn was red?
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