Problem 46 - Algebra and Trigonometry

Question

\(45-46\) A card is drawn at random from a standard 52 -card deck. Determine whether the events \(E\) and \(F\) are mutually exclusive. Then find the probability of the event \(E \cup F .\) (a) \(E :\) The card is a club. F: The card is a king. (b) \(E :\) The card is an ace. F: The card is a spade.

Step-by-Step Solution

Answer

Both Event sets are not mutually exclusive, and the probability \( P(E \cup F) \) is \( \frac{4}{13} \) for both cases.

Step 1: Define Mutually Exclusive Events

Two events, \(E\) and \(F\), are mutually exclusive if they cannot occur at the same time. This means there are no elements in the intersection of \(E\) and \(F\), written as \(E \cap F = \emptyset\).

Step 2: Determine if Events are Mutually Exclusive (Part a)

For part (a), we have:- \(E\): The card is a club.- \(F\): The card is a king.These events are not mutually exclusive because there is one card that is both a club and a king—the King of Clubs. Therefore, \(E \cap F\) is not empty.

Step 3: Determine the Probability of \(E \cup F\) (Part a)

The probability of the union of two events \(E\) and \(F\) can be calculated as:\[ P(E \cup F) = P(E) + P(F) - P(E \cap F) \]- \(P(E)\) = Probability the card is a club = \(\frac{13}{52}\)- \(P(F)\) = Probability the card is a king = \(\frac{4}{52}\)- \(P(E \cap F)\) = Probability the card is the King of Clubs = \(\frac{1}{52}\)Substitute to find:\[ P(E \cup F) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} \]

Step 4: Determine if Events are Mutually Exclusive (Part b)

For part (b), we have:- \(E\): The card is an ace.- \(F\): The card is a spade.The Ace of Spades is both an ace and a spade, so these events are not mutually exclusive. Therefore, \(E \cap F\) is not empty.

Step 5: Determine the Probability of \(E \cup F\) (Part b)

The probability of the union of two events \(E\) and \(F\) can be calculated as:\[ P(E \cup F) = P(E) + P(F) - P(E \cap F) \]- \(P(E)\) = Probability the card is an ace = \(\frac{4}{52}\)- \(P(F)\) = Probability the card is a spade = \(\frac{13}{52}\)- \(P(E \cap F)\) = Probability the card is the Ace of Spades = \(\frac{1}{52}\)Substitute to find:\[ P(E \cup F) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} \]

Key Concepts

Mutually Exclusive Events

When studying probability, you will often come across the concept of "mutually exclusive events." This is a vital concept that helps in determining the likelihood of multiple events happening at the same time.
Simply put, two events are mutually exclusive if they cannot both occur simultaneously. For instance, when flipping a coin, you can't get both heads and tails at once from a single flip. If events are mutually exclusive, the intersection of these events is zero, denoted as \( E \cap F = \emptyset \). This means there are no outcomes common to both events.
In our exercise, we explored two scenarios: whether drawing a card from a deck results in mutually exclusive events or not. For example, drawing a King of Clubs falls into both the categories of clubs and kings, indicating that these two events are not mutually exclusive. Understanding whether events are mutually exclusive is critical for calculating the probabilities of complex outcomes.

Standard Deck of Cards

A standard deck of cards is often used to explore probability problems due to its structured organization and easily quantifiable outcomes.
This deck consists of 52 cards, which are split into four suits: clubs, diamonds, hearts, and spades. In each suit, there are 13 cards comprising numbers 2 through 10, and the face cards: Jack, Queen, King, and Ace.

• Each suit has exactly one of each face card and one of each number card.
• There are 4 cards of each type across the suits, such as 4 Aces or 4 Kings in total.

Understanding this breakdown is essential when calculating probabilities. For example, the chance of drawing a club from a full deck is \( \frac{13}{52} \), given there are 13 clubs in the deck. It's important to have this deck composition in mind when tackling probability problems involving cards.

Union of Events

In probability theory, the union of events is a fundamental notion that describes the event that either one or several events occur.
The union, denoted as \( E \cup F \), refers to the probability that at least one of the events occurs. To calculate this, you need to use the formula:
\[ P(E \cup F) = P(E) + P(F) - P(E \cap F) \]
This formula accounts for the fact that if two events are not mutually exclusive, their overlap (or intersection) might be counted twice if simply added together. By subtracting \( P(E \cap F) \), we adjust for this overlap.

• Example: In part (a) of our exercise, the chance of drawing either a club or a king is calculated by adding their individual probabilities and subtracting the overlap's probability — the likelihood of drawing the King of Clubs.

Using this approach ensures you correctly account for all possible outcomes without double-counting any shared outcomes.