Question 1183890
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.