Conditional probability of first ball being blue given that the second ball is red

Start with an urn with 5 red and 3 blue balls in it. Draw one ball. Put that ball back in the urn along with another ball of the same color. Now draw another ball from the urn.Suppose the second ball is red. What is the probability the first ball was blue.

The answer is 1/3, but why can't I use Bayes Theorem for this?

P(D1 = B | D2 = R) = P(D2 = R | D1 = B) * P(D1 = B) / P(D2 = R) = 5/10 * 3/8 / (5/8) = 3/10

$$P(D_2=R) = P(D_2=R|D_1=B)\times P(D_1=B) + P(D_2=R|D_1=R)\times P(D_1=R)$$
$$P(D_2=R)=\frac{5}{9} \times \frac{3}{8}+\frac{6}{9} \times \frac{5}{8}= \frac{5}{8}$$
Now, $$P(D_1=B|D_2=R) = \frac{P(D_1=B,D_2=R)}{P(D_2=R)} = \frac{\frac{3}{8}\frac{5}{9}}{\frac{5}{8}} = \frac{1}{3}$$
• Exactly how did you determine those fractions? The presence of "10" in the denominators is particularly mysterious, given that in the second draw the urn contains only nine balls. And what does this probability have to do with the chance that the first ball is blue, which I presume you would write as "$P(D_1=B)$"? – whuber May 5 '15 at 14:50
here, you are required to Put that ball back in the urn along with another ball of the same color