Probability help? Jar A has 4 red and 5 black candies. Jar B has 6 red and 2 black candies...

Jar A has 4 red and 5 black candies. Jar B has 6 red and 2 black candies. A fair die is rolled and jar A is selected if a number divisible by 3 comes up, otherwise, Jar B is selected. One candy is drawn from the jar.

I got 'a', 'b' and 'c' but can someone help me with 'd' and 'e'. i don't know how to do them... :(

a) What is the probability you selected Jar A and got a red candy?
b) What is the probability you selected Jar B and got a red candy?
c) What is the probability you got a red candy?
d) Suppose a red candy is drawn, what is the probability it came from jar A?
e) What is the probability Jar B was selected if a black candy is drawn?

a. 2/6 *4/9 = 8/54
b. 4/6 *6/8 = 1/2 = 27/54
c. 35/54
deliberately kept unreduced fractions to simplify further calculations
d. P[jar A | red] = (8/54)/(35/54) = 8/35 <--------

e. P[jar B & black] = 4/6 * 2/8 = 1/6 , P[black] = 1 -35/54 = 19/54,
P[jar B | black] = (1/6)/(19/54) = 9/19 <-------

a) p = 2/6 x 4/9 = 8/54 = .148148 or 14.81%
b) p = 4/6 x 6/8 = 24/48 = 1/2 = .5 or 50%
c) p = 8/54 + 1/2 = 35/54 = .648148 or 64.81%
d) p = .148148/.648148 = .22857 or 22.86% (a/c)
e) p_Black if B = 4/6 x 2/8 = 8/48 = 1/6 = .1667 or 16.67%
__p_Black if A = 2/6 x 5/9 = 10/54 = .185185 or 18.52%
__p_B if Black = .1666/(.1666 + .185185) = .4737 or 47.37%
(The p_B if Black is called "posterior probability")