# A card is drawn from a shuffled deck of 52 cards, and not replaced. Then a second card is drawn. What is the probability that the second card is a king?

Dec 8, 2017

#### Answer:

$\frac{1}{13}$

#### Explanation:

To give a more explained solution of this problem, there are 2 cases you have to consider:

Case 1: The first card drawn is a king
Case 2: The first card drawn is not a king

The reason there's a difference is because in Case 1 the taking of a king on the first card means there is a smaller chance of getting a king on the second card (because the originally taken card is not replaced).

To get the probability of the 2nd card being a king, we can find each individual probability for Cases 1 and 2 and add them together since each of those possibilities are disjoint; in other words, it's not possible that the first card drawn is a king and not a king at the same time.

Case 1

If the first card drawn is a king, the probability of that happening is $\frac{4}{52} = \frac{1}{13}$. The probability of the 2nd card being a king as well would then be $\frac{3}{51}$, since there is one less king possible to be drawn. Multiplying these together gives us: $\frac{1}{13} \cdot \frac{3}{51} = \frac{3}{663} = \frac{1}{221}$

Case 2

If the first card drawn is not a king, the probability of that happening is $\frac{48}{52} = \frac{12}{13}$. (This is because there are 48 cards we're interested in which aren't kings, out of 52 total cards). The 2nd card being a king has a probability of $\frac{4}{51}$, since all 4 kings are still available out of 51 cards left in the deck. Multiplying these together gives us: $\frac{12}{13} \cdot \frac{4}{51} = \frac{48}{663} = \frac{16}{221}$.

Answer

Adding these 2 possibilities together gives the overall probability of drawing a king on the 2nd draw: $\frac{1}{221} + \frac{16}{221} = \frac{17}{221} = \frac{1}{13}$