Dis-Card

Given that Jim picked an ace as his third card, lets find what is the probability that the two other cards are both NOT an ace. So now there are 11 cards in the pack and two of them are aces (the third is the ace Jim has chosen). The probability that the two other cards are not aces is the probability that the firs of them is not an ace, times the probability that the second one is not an ace given that the first one isn't an ace - that's a use of conditional probability .

$P(first \ card \ is \ not \ an \ ace)=\frac{num\ of\ non\ aces\ in\ deck}{num\ of\ all\ cards\ in\ deck}=\frac{9}{11}$

$P(second\ card\ is\ not\ an\ ace\ |\ first\ card\ is\ not\ an\ ace)=\frac{num\ of\ non\ aces\ in\ deck}{num\ of\ all\ cards\ in\ deck}=\frac{8}{10}$

$P(first\ card\ is\ not\ an\ ace\ \cap\ second\ card\ is\ not\ an\ ace)\\ =P(first \ card \ is \ not \ an \ ace) \times P(second\ card\ is\ not\ an\ ace\ |\ first\ card\ is\ not\ an\ ace)\\ =\frac{9}{11}\times \frac{8}{10}=\frac{36}{55}$

Now, the situation where at least one discarded card was an ace is the complementary event of the event who's probability we've just calculated. Therefore, it's probability is $1-\frac{36}{55}=\frac{19}{55}=\frac{m}{n}$ , thus the final solution is $m+n=19+55=74$ .

Let $A$ be the event that Jim discarded at least one Ace and $B$ be the event that Jim's third card was an Ace. We want to find $P(A|B),$ which can be found using the formula

$P(A|B) = \dfrac{P(A \cap B)}{P(B)}.$

To find the value of this fraction, we find the probability of him choosing an Ace for his third card based on how many Aces he discarded initially:

Case 1 : Jim discarded no Aces.

Let $P_0$ be the probability that Jim discarded no Aces and drew an Ace for his third card. If Jim discarded no Aces, then both cards he discarded must have came from the 9 non-Ace cards. There are $\dbinom{9}{2}$ ways to choose the cards. Since there are $\dbinom{12}{2}$ ways to choose two cards from a deck of 12, the probability that Jim discarded no Aces is equal to $\dfrac{\binom{9}{2}}{\binom{12}{2}}.$ When Jim chooses the third card, all three Aces are still there, making the probability that he chooses an Ace for his third card $\dfrac{3}{10}.$ Therefore,

$P_0 = \dfrac{\binom{9}{2}}{\binom{12}{2}} \cdot \dfrac{3}{10} = \dfrac{12}{22} \cdot \dfrac{3}{10}.$

Case 2 : Jim discarded exactly one Ace.

Let $P_1$ be the probability that Jim discarded exactly one Ace and drew an Ace for his third card. If Jim discarded exactly one Ace, then he must have chosen one card out of the three Aces and one card out of the nine other cards. There are $\dbinom{3}{1}\dbinom{9}{1}$ ways to do this, and the probability that Jim discarded exactly one Ace is equal to $\dfrac{\binom{3}{1}\binom{9}{1}}{\binom{12}{2}}.$ Since Jim removed one Ace already, there are only two Aces left in the deck of 10 cards to choose from, making the probability that he chooses an Ace for his third card $\dfrac{2}{10}.$ Therefore,

$P_1 = \dfrac{\binom{3}{1}\binom{9}{1}}{\binom{12}{2}} \cdot \dfrac{2}{10} = \dfrac{9}{22} \cdot \dfrac{2}{10}.$

Case 3 : Jim discarded two Aces.

Let $P_2$ be the probability that Jim discarded two Aces and drew an Ace for his third card. There are $\dbinom{3}{2}$ ways to choose two Aces from the three in the original deck, so the probability that Jim discarded two Aces is equal to $\dfrac{\binom{3}{2}}{\binom{12}{2}}.$ Since Jim removed two Aces, there is only one Ace left among the ten remaining cards, making the probability that he chooses an Ace for his third card $\dfrac{1}{10}.$ Therefore,

$P_2 = \dfrac{\binom{3}{2}}{\binom{12}{2}} \cdot \dfrac{1}{10} = \dfrac{1}{22} \cdot \dfrac{1}{10}.$

Now, note that event $B$ encompasses all three cases, while $A \cap B$ encompasses only cases 2 and 3. Thus,

$\begin{aligned} P(A|B) &= \dfrac{P(A \cap B)}{P(B)} \\ &= \dfrac{P_1 + P_2}{P_0 + P_1 + P_2} \\ &= \dfrac{\frac{9}{22} \cdot \frac{2}{10} + \frac{1}{22} \cdot \frac{1}{10}}{\frac{12}{22} \cdot \frac{3}{10} + \frac{9}{22} \cdot \frac{2}{10} + \frac{1}{22} \cdot \frac{1}{10}} \\ &= \dfrac{18 + 1}{36 + 18 + 1} \\ &= \dfrac{19}{55}, \end{aligned}$

and $m + n = 19 + 55 = \boxed{74}.$