From a deck of $52$ cards, we draw $5$ cards without returning. Determine the probability that we get at least $4$ face cards (w if we know that we have an ace and no hearts.
$A_4 =$ we get $4$ face cards (at least one of them is an ace) and $1$ card that is not a figure; there is no heart among all those cards
- $A_{4.1} = $ we get $1$ ace, $3$ different face cards and $1$ card that is not a figure; there is no heart among all those cards
- $A_{4.2} = $ we get $2$ aces, $2$ different face cards and $1$ card that is not a figure; there is no heart among all those cards
- $A_{4.3} = $ we get $3$ aces, $1$ different face card and $1$ card that is not a figure; there is no heart among all those cards
$A_5 =$ we get $5$ face cards (at least one of them is an ace); there is no heart among all those cards
- $A_{5.1} = $ we get $1$ ace, $4$ different face cards; there is no heart among all those cards
- $A_{5.2} = $ we get $2$ aces, $3$ different face cards; there is no heart among all those cards
- $A_{5.3} = $ we get $3$ aces, $2$ different face cards; there is no heart among all those cards
$B = $ we get at least $1$ ace and no heart
- $B_{1} = $ we get $1$ ace and $4$ different cards; there is no heart among all those cards
- $B_{2} = $ we get $2$ aces and $3$ different cards; there is no heart among all those cards
- $B_{3} = $ we get $3$ aces and $2$ different cards; there is no heart among all those cards
From conditional probability we can write that:
$$P(A|B) = \frac{P(A)}{P(B)} = \frac{P(A_{4.1}) + P(A_{4.2}) + P(A_{4.3}) + P(A_{5.1}) + P(A_{5.2}) + P(A_{5.3})}{P(B_1) + P(B_2) + P(B_3)}$$
I think that:
- $A_{4.1} = {{3}\choose{1}} {{3 \cdot 3}\choose{3}} {{3 \cdot 9}\choose{1}} = 3 \cdot {{9}\choose{3}} \cdot 27 = 3 \cdot \frac{9!}{3! \cdot 6!} \cdot 27 = 9 \cdot 4 \cdot 7 \cdot 27$
- $A_{4.2} = {{3}\choose{2}} {{3 \cdot 3}\choose{2}} {{3 \cdot 9}\choose{1}} = {{3}\choose{2}} {{9}\choose{2}} \cdot 27 = 3 \cdot 9 \cdot 4 \cdot 27$
- $A_{4.3} = {{3}\choose{3}} {{3 \cdot 3}\choose{1}} {{3 \cdot 9}\choose{1}} = 9 \cdot 27$
- $A_{5.1} = {{3}\choose{1}} {{3 \cdot 3}\choose{4}} = 3 \cdot {{9}\choose{4}} = 9 \cdot 7 \cdot 6$
- $A_{5.2} = {{3}\choose{2}} {{3 \cdot 3}\choose{3}} = {{3}\choose{2}} \cdot {{9}\choose{3}} = 9 \cdot 4 \cdot 7$
- $A_{5.3} = {{3}\choose{3}} {{3 \cdot 3}\choose{2}} = {{9}\choose{2}} = 9 \cdot 4$
- $B_{1} = {{3}\choose{1}} {{3 \cdot 3}\choose{4}} = 3 {{9}\choose{4}} = 9 \cdot 7 \cdot 6$
- $B_{2} = {{3}\choose{2}} {{3 \cdot 3}\choose{3}} = {{3}\choose{2}} {{9}\choose{3}} = 9 \cdot 4 \cdot 7$
- $B_{3} = {{3}\choose{3}} {{3 \cdot 3}\choose{2}} = {{9}\choose{2}} = 9 \cdot 4$
So we have that:
$$P(A|B) = \frac{9 \cdot 4 \cdot 7 \cdot 27 + 3 \cdot 9 \cdot 4 \cdot 27 + 9 \cdot 27 + 9 \cdot 7 \cdot 6 + 9 \cdot 4 \cdot 7 + 9 \cdot 4}{9 \cdot 7 \cdot 6 + 9 \cdot 4 \cdot 7 + 9 \cdot 4} = \frac{4 \cdot 7 \cdot 27 + 3 \cdot 4 \cdot 27 + 27 + 7 \cdot 6 + 4 \cdot 7 + 4}{ 7 \cdot 6 + 4 \cdot 7 + 4} = \frac{756 + 324 + 27 + 42 + 28 + 4}{42 + 28 + 4} = \frac{1181}{74}$$
The answer is obviously wrong. I don't know where is the error in my reasoning.
I assume you intend a "figure" card to be one whose rank is J, Q, K, or A.
Under that interpretation, most of your calculations make sense up until you get to calculating the counts for your $B$ events. $B_1$ for instance should have been the count of the number of possible hands with exactly once ace and no hearts in hand. You choose which ace it was, and then from all remaining non-heart cards (of which there are many) you choose four additional cards to make up the hand.
You erroneously used $\binom{3\cdot 3}{4}$ here to count how many ways there were to finish out the hand, but that is the count had the cards to finish out the hand all been face cards. We are not limiting ourselves to face cards here. There are $12$ ranks remaining which are not aces for a total of $3\cdot 12$ non-ace non-heart cards to choose from, many more than the $3\cdot 3$ that you accidentally limited yourself to.
Using $3\cdot 12$ here in those calculations for the $B$'s should correct the mistake.