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Not sure how to set these up correctly.. Any help is appreciated, thank you!

You draw 5 cards from a 52-card deck. Find the probability of each scenario.

  1. Two cards are sevens, given there are no face cards
  2. There is exactly one face card, given three of the cards are red
  3. Three cards are Aces, given one of the Aces was removed from the deck
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closed as off-topic by barak manos, Parcly Taxel, Watson, Shailesh, Ethan Bolker Nov 2 '16 at 14:03

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  • $\begingroup$ What exactly do you mean by "set up"? $\endgroup$ – Graham Kemp Nov 2 '16 at 5:36
  • $\begingroup$ @GrahamKemp Presumably (and I've seen this quite a lot) this means "apply known techniques to the problem at hand so as to reduce it to a bunch of computations". Which only comes with practice. $\endgroup$ – Parcly Taxel Nov 2 '16 at 5:52
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    $\begingroup$ You could by the least specify how many face cards exist in a deck. You seem to be making ZERO effort here, other than copy-pasting your homework assignment, so be advised that this is not a 'do my homework for free' service! $\endgroup$ – barak manos Nov 2 '16 at 6:02
  • $\begingroup$ Please check what I have so far below $\endgroup$ – Victoria Nov 2 '16 at 7:32
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The probability of drawing $k_1$ from $n_1$ category-1 cards, $k_2$ from $n_2$ category-2 cards, ..., and $k_r$ from $n_r$ category-$r$ cards, when given that you are drawing $k_1+k_2+\cdots+k_r$ cards from all $r$ categories is

$$\dfrac{\dbinom{n_1}{k_1}\dbinom{n_2}{k_2}\cdots\dbinom{n_r}{k_r}}{\dbinom{n_1+n_2+\cdots +n_r}{k_1+k_2+\cdots+ k_r}}$$

All you have to do is identify how many from what many of which categories you are drawing.

(You may also need to employ Bayes' Theorem.)


  1. Two cards are sevens, given there are no face cards

The probability of drawing 2 from 4 sevens and 3 from 36 non-face-nor-sevens when drawing 5 from 40 non-face cards is:

$$\dfrac{\binom 4 2\binom {36}3}{\binom{40}5}$$


  1. There is exactly one face card, given three of the cards are red

The probability of drawing (I) 1 red face, 2 red non-face, 2 black non-face, or (II) 1 black face, 3 red non-face, 1 black non-face, when drawing 3 red and 2 black cards from the deck is:

$$\frac{\binom 6 1\binom {20}2\binom{20}2+\binom 6 1\binom {20}3\binom{20}1}{\binom{26}3\binom{26}2}$$


  1. Three cards are Aces, given one of the Aces was removed from the deck

The probability of drawing 3 from 3 aces, 2 from 48 non-aces when drawing 5 from 51 remaining cards is:

$$\dfrac{\binom 33\binom {48}2}{\binom{51}5}$$


Remark: When to multiply and when to add.   By the universal principle of counting, the count of ways to perform a sequence of tasks is the product of ways to perform the tasks, while the count of ways to perform alternative tasks is the sum of ways to perform the alternatives.

In short: Multiply the counts of "and" joined tasks, else sum the counts of "or" joined tasks.

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  • $\begingroup$ So I'm reading the first question as drawing exactly 2 seven cards and none of them can be face cards. There are 12 face cards and 4 sevens. $\binom{12}{0}$ drawing no face cards, $\binom{4}{2}$ drawing 2 seven cards, and $\binom{36}{3}$ drawing 3 additional cards from the remaining deck. I got 36 from subtracting the total number of face cards and sevens from 52. $$\frac{\binom{12}{0}+\binom{4}{2}+\binom{36}{3}} {\binom{52}{5}}$$ Am I on the right track? $\endgroup$ – Victoria Nov 2 '16 at 7:05
  • $\begingroup$ @Victoria, no; not at all. Why are you adding? Also, you wish the probability of drawing $2$ from $4$ sevens and $3$ from $36$ non-face-nor-sevens when given that you are drawing $5$ from $40$ non-face cards. $\endgroup$ – Graham Kemp Nov 2 '16 at 7:52
  • $\begingroup$ So it's simply $$\frac{\binom{4}{2}\binom{36}{3}}{\binom{40}{5}}$$ ? $\endgroup$ – Victoria Nov 2 '16 at 9:20
  • $\begingroup$ Thank you for your patience by the way. Also for the 2nd question I have set up picking 1 face card $\binom{12}{1}$ , 3 red cards $\binom{26}{3}$, and 1 additional card $\binom{37}{1}$ giving me $$\frac{\binom{12}{1}\binom{26}{3}\binom{37}{1}}{\binom{52}{5}}$$ $\endgroup$ – Victoria Nov 2 '16 at 9:46
  • $\begingroup$ @Victoria No. This one is where you use addition. The face card may be red or black, so there are two (disjoint) paths. (A) Select $1$ red face, $2$ red non-face, and $2$ black non-face cards OR (B) Select $3$ red non-face, $1$ black face, and $1$ black non-face; given that you are selecting $3$ red and $2$ black cards from the deck.$$\frac{\binom{\Box}{1}\binom{\Box}{2}\binom{\Box}{2}+\binom{\Box}{3}\binom{\Box}{1}\binom{\Box}{1}}{\binom{\Box}{3}\binom{\Box}{2}}$$ $\endgroup$ – Graham Kemp Nov 2 '16 at 11:36
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First off, you should try to make your question a bit more explicit, even though the text you were given might be a bit obscure itself.
I will give you some hints, but I will not formalize it, so you will have to "set up" things by yourself.

Let's begin with the third scenario.
In this case, you remove an ace from the deck, so you modify it before the event. You can visualize the same problem but with a new deck, so the question scenario would become "Five cards are drawn from a 52-card deck where an ace has been remove, and three of the five cards drawn are aces".
This is a simple scenario with no conditional probability issue.

On the second scenario, three of the five cards drawn are red. Does it influence the probability that exactly one card is a face? Is a red card more likely to be a face than a black card?

I do not really see some hints to give you on the first scenario. However, keep in mind that "The probability of X, given Y" is exactly equivalent to $P(X|Y)$. Which means, in a dream world where Y has already occurred and you cannot do anything to it, the probability that X occurs.
It might sound obvious, but you can quickly become lost when working with probabilities, so it could be good to keep this in mind.

Good luck and don't give up, maths are cool!

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  • $\begingroup$ The third one was the only one I had somewhat of an idea as to what to do but the "given one of the Aces was removed from the deck" statements confuses me a bit. Does that mean I am now picking 5 cards from a 51 card deck? If so I came up with this set up: $\binom{3}{3}$ picking three aces out of the three in the deck since one was removed. $\binom{48}{2}$ picking 2 additional cards from the deck. $$\frac{\binom{3}{3} + \binom{48}{2}} {\binom{53}{48}}$$ $\endgroup$ – Victoria Nov 2 '16 at 6:27
  • $\begingroup$ I would say an accurate reasoning is as follows: $\endgroup$ – Right Leg Nov 2 '16 at 6:34
  • $\begingroup$ I would suggest such a reasoning: 1 - Pick an ace: $3/51$ 2 - Pick an ace: $2/50$ 3 - Pick an ace: $1/49$ 4 - Pick a card that is not an ace: $48/48$ 5 - Pick a card that is not an ace: $47/47$ You can try these operations in a different order, it should lead to the same result. For instance: 1 - Pick a card that is not an ace: $48/51$ 2 - Pick a card that is not an ace: $47/50$ ... $\endgroup$ – Right Leg Nov 2 '16 at 6:39
  • $\begingroup$ @Victoria The probability of picking $3$ from the $3$ remaining aces, and $2$ from the $48$ other cards, when picking $5$ cards from the $51$ left in the deck is: $$\frac{\binom{3}{3}\binom{48}{2}}{\binom{51}{5}}~=~\text{what Right Leg suggests}$$ Note: Multiplication rather than addition, and I have no idea where you plucked the numbers for the denominator's binomial coefficient. $\endgroup$ – Graham Kemp Nov 2 '16 at 8:01

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