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You are dealt five cards from a standard and shuffled deck of playing cards. Note that a standard deck has 52 cards and four of those are kings. What is the probability that you'll have at most three kings in your hand?

I know that the answer is $\frac{54144}{54145}$ from the answer key, and I know that the sample space is ${^{52}\mathrm C_5}$. What I don't get is how to find the event.

Do I just add the combination for each number of kings together (${^5\mathrm C_2} + {^5\mathrm C_3}$)?

Or do I need to multiply as well to account for the other cards in the 5-card hand?

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Formatting tips here. – probablyme Mar 16 at 4:07
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Also, consider giving a check mark. – probablyme Mar 16 at 4:10

The probability that you will have at most 3 kings is the probability that you will have less than 4.

$$\mathsf P(K\leq 3) = 1 -\mathsf P(K=4)$$

The probability that you will have exactly all four kings is the count of ways to select 4 kings and 1 other card divided by the count of ways to select any 5 cards.

$$\mathsf P(K=4)~=~\dfrac{{^{4}\mathrm C_{4}}\cdot{^{(52-4)}\mathrm C_{(5-4)}}}{^{52}\mathrm C_5}$$

Put it together.

$$\mathsf P(K\leq 3) ~=~ 1 -\dfrac{{^{4}\mathrm C_{4}}\cdot{^{(52-4)}\mathrm C_{(5-4)}}}{^{52}\mathrm C_5}$$

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This does not agree with the other answers when calculated. It should be a $\binom{48}{1}$ instead of a $\binom{48}{4}$, as you stated in your explanation. – Paddling Ghost Mar 16 at 4:25
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Typo,.... corrected. Thank you. – Graham Kemp Mar 16 at 4:46
    
@PaddlingGhost Ah, I see what you meant. Good eye. I thought I read a one, my mistake. – probablyme Mar 16 at 4:54

The number of ways to select $5$ cards that have $0,1,2,3$ kings are: $ \binom{48}{5}, \binom{4}{1}\binom{48}{4}, \binom{4}{2}\binom{48}{3}, \binom{4}{3}\binom{48}{2}$ respectively. Thus the probability of at most $3$ kings is: $Pr(x \leq 3) =\dfrac{\binom{48}{5}+\binom{4}{1}\binom{48}{4}+\binom{4}{2}\binom{48}{3}+\binom{4}{3}\binom{48}{2}}{\binom{52}{5}}$

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At most 3 kings is the same thing as not 4 kings.

$P(4$ kings)$ = 48 / {52\choose 5} = \frac{1}{54145}$

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This is wrong. Please review the other responses. – probablyme Mar 16 at 4:04
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@probablyme I believe you are mistaken. The Graham Kemp's answer has an error. – Paddling Ghost Mar 16 at 4:22
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You're right, but it's easy to misread $P(4 \text{ kings})=\frac{1}{54145}$ as saying that the final answer should be $\frac{1}{54145}$. – Kevin Mar 16 at 5:05
    
@Kevin I agree with you. And not just because of your awesome name – Kevin Mar 16 at 14:31
    
@Kevin Since the final answer was known in advance, I do not think misreading is so likely. – Jeppe Stig Nielsen Mar 16 at 19:08

Yet another way: there are 52 cards in the deck, 4 of which are kings. The probability the first card is a king is 4/52= 1/13. There are then 51 cards left, 3 of them kings. The probability the second card is a king is 3/51= 1/17. There are then 50 cards left, 2 of them kings. The probability the third card is a king is 2/50= 1/25. There are then 49 cards left, 1 of them a king. The probability the fourth card is a king is 1/49. There are no longer any kings in the deck so the probability the fifth card is not a king is 1.

The probability of getting "four kings and one non-king" in that order is (1/13)(1/17)(1/25)(1/49). There are 5 ways to order "four kings and one non- king" (the non-king being in any of the 5 places) so the probability of "four kings and one non-king" in any order is 5(1/13)(1/17)(1/25)(1/49).

Finally, the probability of get "at most three kings in a five card hand" is 1- 5(1/13)(1/17)(1/25)(1/49).

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Since you need to find probability $P(\leq 3~\text{kings})$, you can find $P(>3~\text{kings})$ and subtract it from $1$.

$P(>3~\text{kings}) \iff P(4~\text{kings})$, which is $48/52C5 = 1/54145$

Here, we take $48$, since out of $5$ cards four are kings, and fifth can be any one out of the remaining $48$ cards.

So your final answer is $1 - (1/54145)$ which is $54144/54145$.

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Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Mar 16 at 10:27

Split it into disjoint events, and then add up their probabilities:


The probability of exactly $\color\red0$ kings is:

$$\frac{\binom{4}{\color\red0}\cdot\binom{52-4}{5-\color\red0}}{\binom{52}{5}}$$


The probability of exactly $\color\red1$ king is:

$$\frac{\binom{4}{\color\red1}\cdot\binom{52-4}{5-\color\red1}}{\binom{52}{5}}$$


The probability of exactly $\color\red2$ kings is:

$$\frac{\binom{4}{\color\red2}\cdot\binom{52-4}{5-\color\red2}}{\binom{52}{5}}$$


The probability of exactly $\color\red3$ kings is:

$$\frac{\binom{4}{\color\red3}\cdot\binom{52-4}{5-\color\red3}}{\binom{52}{5}}$$


Hence the overall probability is:

$$\sum\limits_{n=0}^{3}\frac{\binom{4}{n}\cdot\binom{52-4}{5-n}}{\binom{52}{5}}=\frac{54144}{54145}$$

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