# From a deck of $40$ cards with four kings, we extract at random $5$ cards. What is the probability of drawing $1$, and only $1$, king?

Question:

From a deck of $40$ cards we extract, at random, $5$ cards. What is the probability of drawing $1$, and only $1$, king?

The deck has $4$ suits - $Hearts \color{red}{\heartsuit}$, $Diamonds \color{red}{\diamondsuit}$, $Clubs \clubsuit$ and $Spades \spadesuit$ - and each of these has cards numbered from $2$ to $7$, an $Ace$, a $King$, $Queen$ and $Jack$.

What follows is my approach to the problem:

There are $4$ kings in our deck $$K\spadesuit, K\color{red}{\heartsuit}, K\color{red}{\diamondsuit}, K\clubsuit$$

and $36$ other cards from our deck that are not Kings. We have a total of $$\binom{40}{5}$$ possible outcomes.

Then the probability of, drawing 5 cards at random from the deck, drawing $1$, and only $1$, king is $$\frac{4 \binom{36}{5}}{\binom{40}{5}} \approx 36\%$$ $\tag*{$\blacksquare$}$

Since you draw five cards, after you've counted the ways to choose one king, you count the ways to choose four other cards. In other words $$\frac{4\binom{36}{\color{red}{4}}}{\binom{40}{5}}\approx 0.358$$

$$\frac{\binom{4}{1}\cdot\binom{40-4}{5-1}}{\binom{40}{5}}$$

In addition to the excellent answer from @Max I wanted to add a technique I used to employ when I was taking a course that gave questions of this variety and I did not have access to help and needed to further convince myself that my answer was correct.

Recall it must be the case that the probability of the sample space is 1. So if the probability of an event $A$ is $x_1$ and the probability of the event $\neg A$ is $x_2$, it must be that $x_1$ + $x_2$ = 1. If they don't sum to 1 there is an issue somewhere. Even if they do sum to 1, there may still be an issue, but if they don't you know for certain there is a problem.

In our case the event $A$ is choosing exactly 1 king, so $\neg A$ is the event of choosing 0, 2, 3 or 4 kings. We find the probability of $\neg A$ in the same way as @Max suggested.

$$\frac{\binom{4}{0} \cdot \binom{36}{5} + \binom{4}{2}\cdot\binom{36}{3} + \binom{4}{3} \cdot \binom{36}{2} + \binom{4}{4} \cdot \binom{36}{1}}{\binom{40}{5}} \approx .6419.$$

When summed with the exact value given in the answer by @Max, it becomes the probability of the sample space, and you can check it is exactly equal to 1. When summed with the value you found, it is much higher ($\approx 2.9336$), which implies there was an error in the calculation of either $A$, $\neg A$ or both.

Lastly, I think there was an error in your calculation that

$$\frac{4 \cdot \binom{36}{5}}{\binom{40}{5}} \approx .36$$

because

$$\frac{4 \cdot \binom{36}{5}}{\binom{40}{5}} = \frac{20944}{9139} \approx 2.2917$$

which also indicates an error because it must be the case that probability of an event is less than or equal to 1.

Hopefully what I have provided does not further confuse you on this topic, just some things to consider that I found useful for identifying errors.