A stack of cards consists of six red and five blue cards. A second stack of cards consists of nine red cards. A stack is selected at random and three of its cards are drawn. If all of them are red, what is the probability that the first stack was selected?
let X be the event of drawing three red cards, A be the first deck and B the second.
then P(A) = P(B) = .5
P(X|A) = (6/11 * 5/10 * 4/9) = .121212
P(X|B) = 1
The Attempt at a Solution
P(X|A)P(A) / [P(X|A)P(A) + P(X|B)(B)] = (.1212 * .5) / [(.1212 * .5) + (1 * .5] = .195
I think I did this right. Can anyone confirm or give me a hint as to what might be wrong? Intuitively the 1/5 answer bothers me since it seems to me that the chance of choosing one of the two decks seems like it would still be 1/2 since drawing three red cards does not imply one deck or the other, but the chapter is on Bayes' Formula and not on independence.