# Random Card Probability

I am trying to master probability, but the book I am using does not provide solutions to its exercises. I am confused with b and d, but would like all of my work looked over to make sure I am on the right track. Thank you for your help.

You are randomly dealt a $$5$$ card hand from a set of $$52$$ cards ($$4$$ suits with $$13$$ cards in each suit, $$26$$ red and $$26$$ black).

(a) What is the probability that the suit of the first card you get is Hearts?

Is it simply $$13/52 = 25\%$$?

(b) What is the probability you get $$3$$ red cards and $$2$$ black cards?

$$\frac{\dfrac{26}{52} \cdot \dfrac{25}{51} \cdot \dfrac{24}{50} + \dfrac{26}{52} + \dfrac{25}{51}}{\dbinom{52}{5}}$$ I am unsure if I needed to multiply an $$\binom{n}{k}$$ combination.

(c) What is the probability that you get all $$4$$ aces?

Since the $$4$$ aces will make up $$4$$ of the $$5$$ cards in the hand, there are only $$48$$ other cards to choose for your $$5$$th card. $$\frac{48}{\dbinom{52}{5}}$$

(d) What is the probability that you get exactly one Ace?

I think this is how to do the problem, but I can’t justify it. I know that with exactly one ace, you will have $$48$$ cards to choose for your next card then $$47$$. I do not know why the division by $$4!$$ is needed. $$\frac{\dfrac{48 \cdot 47 \cdot 46 \cdot 45}{4 \cdot 3 \cdot 2 \cdot 1}}{\dfrac{52!}{47!5!}}$$

• Do you know about the hypergeometric distribution? Oct 15, 2017 at 5:38
• If you want to produce $\binom{n}{k}$, type \binom{n}{k} when you are in math mode (place your text between dollar signs). Please read this tutorial on how to typeset mathematics on this site. Oct 16, 2017 at 19:57
• Please do not vandalize your question by deleting its contents. Oct 16, 2017 at 19:58

Your answers to (a) and (c) are correct, but attempts for (b) and (d) are very confused.

These are problems of drawing w/o replacement, and I'll use a uniform approach for all the problems so that you can see a pattern.

$(a): \dfrac{\binom{13}1}{\binom{52}1}$

$(b): \dfrac{\binom{26}3\binom{26}2}{\binom{52}5}$

$(c): \dfrac{\binom44\binom{48}1}{\binom{52}5}$

$(d): \dfrac{\binom41\binom{48}4}{\binom{52}5}$

You would do well to follow up on the suggestion of Lord Shark the Unknown

What is the probability that the suit of the first card you get is Hearts?

The probability is indeed $$\frac{13}{52} = \frac{1}{4}$$

What is the probability that you get three red cards and two black cards when a five-card hand is dealt?

There are $\binom{52}{5}$ five-card hands.

Since there are $26$ red cards, the number of ways you can receive three of them is $\binom{26}{3}$. Since there are $26$ black cards, the number of ways you can receive two of them is $\binom{26}{2}$. By the Multiplication Principle, the number of ways you can receive three red cards and two black cards is $$\binom{26}{3}\binom{26}{2}$$ Therefore, the desired probability is $$\frac{\dbinom{26}{3}\dbinom{26}{2}}{\dbinom{52}{2}}$$

The Multiplication Principle states that if there are $m$ ways of performing one task and $n$ ways of performing another task independently of the first, both tasks can be performed in $mn$ ways. The Addition Principle states that if two tasks are mutually exclusive and there are $m$ ways of performing one of the tasks and $n$ ways of performing the other, then there are $m + n$ ways of performing one of the tasks.

In this problem, we multiply since it is possible to choose two black cards and three red cards from the deck at the same time.

What is the probability that you get all four aces when you are dealt five cards?

You are correct. There is one way to receive all four aces and $48$ ways to receive another card. Hence, the desired probability is $$\frac{\dbinom{4}{4}\dbinom{48}{1}}{\dbinom{52}{5}} = \frac{48}{\dbinom{52}{5}}$$

What is the probability that you are dealt exactly one ace when you are dealt five cards?

To receive exactly one ace, you must receive one of the four aces and four of the other $48$ cards in the deck. Hence, the desired probability is $$\frac{\dbinom{4}{1}\dbinom{48}{4}}{\dbinom{52}{5}}$$

Your answer was incorrect because you did not account for the number of ways the ace could be selected.

In your calculation, you wrote that the number of ways of selecting four of the $48$ cards in the deck that are not aces is $$\frac{48 \cdot 47 \cdot 46 \cdot 45}{4 \cdot 3 \cdot 2 \cdot 1}$$ which is correct. You asked why we must divide by $4!$. The $4!$ represents the number of orders in which the same four cards can be selected. Notice that $$\binom{48}{4} = \frac{48!}{4!(48 - 4)!} = \frac{48!}{4!44!} = \frac{48 \cdot 47 \cdot 46 \cdot 45 \cdot 44!}{4!44!} = \frac{48 \cdot 47 \cdot 46 \cdot 45}{4 \cdot 3 \cdot 2 \cdot 1}$$ Also, since you counted subsets (unordered selections) in the denominator, you must also count subsets in the numerator.

It is possible to use ordered selections in both the numerator and denominator. In this case, there are $52 \cdot 51 \cdot 50 \cdot 49 \cdot 48$ ordered selections of five cards. There are $4$ ways to choose one of the four aces, $48 \cdot 47 \cdot 46 \cdot 45$ ways to make an ordered selection of four of the cards in the deck that are not aces, and five positions in which the ace could be selected, so the number of favorable ordered selections is $$4 \cdot 48 \cdot 47 \cdot 46 \cdot 45 + 48 \cdot 4 \cdot 47 \cdot 46 \cdot 45 + 48 \cdot 47 \cdot 4 \cdot 46 \cdot 45 + 48 \cdot 47 \cdot 46 \cdot 45 \cdot 4$$ which can be simplified to $$5 \cdot 4 \cdot 48 \cdot 47 \cdot 46 \cdot 45$$ Hence, the probability that you are dealt exactly one ace is $$\frac{5 \cdot 4 \cdot 48 \cdot 47 \cdot 46 \cdot 45}{52 \cdot 51 \cdot 51 \cdot 49 \cdot 48}$$ As you should verify, this is equal to $$\frac{\dbinom{4}{1}\dbinom{48}{4}}{\dbinom{52}{5}}$$ Clearly, it is simpler to solve the problem using combinations.