Setting Up Card Conditional Probability Problem

Question

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

Answer 1

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.)

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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:

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

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2. 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:

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

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3. 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:

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

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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.

Answer 2

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!