How to Set Up Card Probability Problem?

Question

Rosa draws a five-card hand from a 52-card deck. For each scenario, calculate the total possible outcomes.

1. Rosa’s hand has exactly two Jacks
2. Rosa’s hand has at least one face card (Jack, Queen, King)

Answer 1

Rosa's hand has exactly two hands.

For the denominator, select five of the $52$ cards in the deck. This can be done in

$$\binom{52}{5}$$

ways. For the numerator, select two of the four Jacks and three of the other $52 - 4 = 48$ cards. This can be done in

$$\binom{4}{2}\binom{48}{5}$$

ways.

Rosa's hand has at least one face card.

The complementary event of selecting at least one face card is a selection of no face cards. To find the desired probability, find the probability of the complementary event, then subtract your answer from $1$. Since each of the three ranks of face cards (Jack, Queen, King) has four suits (club, diamond, heart, spade), there are $3 \cdot 4 = 12$ face cards. Since there are $12$ face cards, there are $52 - 12 = 40$ cards that are not face cards. Thus, the probability of choosing a five-card hand with no face cards is

$$\frac{\dbinom{40}{5}}{\dbinom{52}{5}}$$

so the probability of obtaining at least one face card is

$$1 - \frac{\dbinom{40}{5}}{\dbinom{52}{5}}$$

Answer 2

To set it up, you need to work out what the probability (or event) space is - what are all the possible outcomes that might happen? - and then work out what the space of "successful" outcomes is - what are all the ways that we would count a success?

So in this case, the full space we're looking at is "all possible hands of five cards from a deck of cards". So how many possible hands are there?

Then for the first question, the "successes" are "hands of five cards where two of the cards are Jacks and the other three aren't". How many are there? Consider how you'll deal with hands containing the same cards in a different order, and make sure you're consistent in the two spaces.