Problem 36 - Algebra and Trigonometry

Question

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

(a) Events are not mutually exclusive; \(P(E \cup F) = \frac{4}{13}\). (b) Events are not mutually exclusive; \(P(E \cup F) = \frac{4}{13}\).

Step 1: Understanding Mutually Exclusive Events

Two events are mutually exclusive if they cannot occur at the same time. For instance, drawing a card that is both red and black is impossible, hence a mutually exclusive scenario.

Step 2: Analyzing Events E and F for Part (a)

For part (a), event \(E\) is drawing a club, and event \(F\) is drawing a king. Since there is one king in each suit, it is possible to draw the king of clubs, meaning events \(E\) and \(F\) are not mutually exclusive.

Step 3: Calculating Probability for Part (a)

To find the probability of \(E \cup F\), we use: \(P(E \cup F) = P(E) + P(F) - P(E \cap F)\). There are 13 clubs, so \(P(E) = \frac{13}{52}\), and 4 kings, so \(P(F) = \frac{4}{52}\). There is 1 king of clubs, so \(P(E \cap F) = \frac{1}{52}\).\\[P(E \cup F) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}\]

Step 4: Analyzing Events E and F for Part (b)

For part (b), event \(E\) is drawing an ace, and event \(F\) is drawing a spade. The ace of spades exists, meaning both events can occur simultaneously, so events \(E\) and \(F\) are not mutually exclusive.

Step 5: Calculating Probability for Part (b)

We use the same formula: \(P(E \cup F) = P(E) + P(F) - P(E \cap F)\). There are 4 aces, so \(P(E) = \frac{4}{52}\), and 13 spades, so \(P(F) = \frac{13}{52}\). There is 1 ace of spades, so \(P(E \cap F) = \frac{1}{52}\).\\[P(E \cup F) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}\]

Key Concepts

Mutually Exclusive Events

Mutually exclusive events are a key concept in probability. They refer to two events that cannot happen at the same time. For example, when drawing a card from a deck, the card cannot be a heart and a club simultaneously. If we have two events, such as drawing a king and drawing a club, these events are not mutually exclusive because there is a king of clubs in the deck. Understanding whether events are mutually exclusive helps in calculating probabilities.

• If events are mutually exclusive, the probability of both occurring is zero.
• If not mutually exclusive, they can occur together, like the king of clubs.

In solving problems involving mutually exclusive events, we adjust our probability calculations accordingly, often using the formula for the union of two events.

Card Probability

Card probability involves calculating the chances of drawing certain cards from a standard 52-card deck. This kind of probability is fundamental in learning how to handle random events. A deck consists of 4 suits: clubs, diamonds, hearts, and spades, each containing 13 cards including one king, one ace, and others.

• Probability of drawing a club: Since there are 13 clubs, it is \( P(E) = \frac{13}{52} \).
• Probability of drawing a king: With 4 kings in the deck, \( P(F) = \frac{4}{52} \).

It's important to understand that drawing a specific card, such as the king of clubs, involves both the suit and the rank, thus affecting the probability calculations. When both events can occur together, it modifies how we calculate total probabilities.

Union of Events

The union of events in probability refers to any situation where any of the events occurs. To find the probability of the union of two events, we use the formula: \[ P(E \cup F) = P(E) + P(F) - P(E \cap F) \]This formula helps us ensure we do not count the same event more than once. For instance, when considering the events of drawing a club or a king:

• Calculate individual probabilities: \( P(E) = \frac{13}{52} \) and \( P(F) = \frac{4}{52} \).
• Subtract the overlap: if there's 1 card that is both a king and a club (King of clubs), \( P(E \cap F) = \frac{1}{52} \).
• Resultant union probability: \( P(E \cup F) = \frac{16}{52} = \frac{4}{13} \).

This approach gives a comprehensive method for calculating combined event probabilities, essential for many real-world scenarios and more complex probability questions.