Question

A card is selected from a shuffled deck. What is the probability that it is either a king or a club? That it is both a king and a club?

Short answer

Probability of a king or club is \( \frac{4}{13} \). Probability of both a king and a club is \( \frac{1}{52} \).

Step-by-step solution

Calculate the Probability of Drawing a King

There are 4 kings in a standard deck of 52 cards. To find the probability of drawing a king, divide the number of kings by the total number of cards: \[ P(\text{King}) = \frac{4}{52} = \frac{1}{13} \]

Calculate the Probability of Drawing a Club

There are 13 clubs in a standard deck of 52 cards. To find the probability of drawing a club, divide the number of clubs by the total number of cards: \[ P(\text{Club}) = \frac{13}{52} = \frac{1}{4} \]

Calculate the Probability of Drawing a King and a Club

There is only 1 card that is both a king and a club (the King of Clubs) in a standard deck. To find this probability, divide the number of King of Clubs cards by the total number of cards: \[ P(\text{King and Club}) = \frac{1}{52} \]

Calculate the Probability of Drawing Either a King or a Club

To find the probability of drawing either a king or a club, use the formula for the union of two events: \[ P(\text{King or Club}) = P(\text{King}) + P(\text{Club}) - P(\text{King and Club}) \] Substitute the values calculated in the previous steps: \[ P(\text{King or Club}) = \frac{1}{13} + \frac{1}{4} - \frac{1}{52} \] Convert all fractions to a common denominator and solve: \[ P(\text{King or Club}) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} \]

Key concepts

Combinatorics

Combinatorics is a branch of mathematics focused on counting, arrangement, and combination of objects. It's widely used in probability theory to determine the number of possible outcomes in a given scenario. For example, when calculating the probability of drawing a king or a club from a deck of cards, we use combinatorics to count the number of kings and clubs.
If we want to determine how many ways we can choose a single card from a deck, we can use the combination formula: \ \[C(n, k) = \frac{n!}{k! (n - k)!}\]Here, \(n\) is the total number of items, and \(k\) is the number of items to choose. In our case, we are choosing 1 card out of 52, so \(C(52, 1) = 52\).

Deck of Cards

A standard deck of cards plays a crucial role in many probability exercises. A standard deck consists of 52 cards divided into four suits: hearts, diamonds, clubs, and spades. Each suit contains 13 cards, ranging from Ace to King.
Knowing these details helps in calculating probabilities related to drawing specific cards:

• There are 4 kings in the deck.
• There are 13 clubs in the deck.
• There is only 1 King of Clubs.

Understanding the composition of the deck allows us to compute probabilities such as drawing a king, a club, or a card that is both a king and a club.

Union of Events

In probability theory, the union of events is used to calculate the probability of any one of several events occurring. When dealing with multiple events, like drawing a king or a club, we use the formula for the union of events: \ \[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]This formula accounts for the overlap between events, ensuring we don't double-count outcomes that satisfy both conditions. In our card example, the union of events helps find the probability of drawing either a king or a club: \ \[P(\text{King or Club}) = \frac{1}{13} + \frac{1}{4} - \frac{1}{52} = \frac{4}{13}\]

Independent Events

Independent events in probability are events whose occurrence or non-occurrence does not affect each other. In other words, the outcome of one event does not influence the outcome of another.
In the context of drawing a card from a deck, if we were to draw one card and then replace it before drawing another, these events would be independent. However, in our exercise, we are only drawing one card, so the events are considered dependent.

• The probability of drawing a king: \ \[\frac{4}{52} = \frac{1}{13}\]
• The probability of drawing a club: \ \[\frac{13}{52} = \frac{1}{4}\]
• The probability of drawing a King of Clubs: \ \[\frac{1}{52}\]

These probabilities help us understand the chances of individual and combined outcomes efficiently.