SOLUTION: Urn 1 contains 2 white, 3 blue, and 4 red balls. Urn 2 contains 3 white and 5 blue balls. A ball is randomly drawn from urn 1, without replacement, and then placed inside urn 2. A

Question 1183890: Urn 1 contains 2 white, 3 blue, and 4 red balls. Urn 2 contains 3 white and 5 blue balls. A ball is randomly drawn from urn 1, without replacement,
and then placed inside urn 2. A ball is then drawn randomly from urn 2 and placed back in urn 1. If a ball is to be randomly drawn from urn 1,
what is the probability that the ball is red? Is there a change in the probability of red from before the experiment? How would you explain your answer?
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
I have solved a similar problem earlier. This other problem is different because urn 1 has extra 4 red balls. The reasoning is similar, though, and
this problem should be EASY, although might take some time in sorting out the appropriate probabilities.

P(wwr) = (2/9)*(4/9)*(4/9) = 32/729
P(wbr) = (2/9)*(5/9)*(4/9) = 40/729
P(bwr) = (3/9)*(3/9)*(4/9) = 36/729
P(bbr) = (3/9)*(6/9)*(4/9) = 72/729
P(rwr) = (4/9)*(3/9)*(3/9) = 36/729
P(rbr) = (4/9)*(5/9)*(3/9) = 60/729
P(rrr) = (4/9)*(1/9)*(4/9) = 16/729

Hence. P(red 2nd time around) = sum of all preceding probabilities = 292/729 ~ 0.40, to 2 d.p.
The probability of 0.40 in drawing red the 2nd time around is LOWER than the probability of red (4/9 or 0.44 to 2 d.p.) in the first draw.

The lower probability comes from the fact that the first draw could be a red, but that same red may not be picked from the 2nd urn,
thereby lowering the number of red balls in urn 1.
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