Problem 22 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

A card is drawn from a well-shuffled deck of 52 playing cards. Let \(E\) denote the event that the card drawn is an ace and let \(F\) denote the event that the card drawn is a diamond. Determine whether \(E\) and \(F\) are independent events. Give an intuitive explanation for your answer.

Step-by-Step Solution

Answer

Events E and F are independent because the probability of both events occurring together, \(P(E \cap F) = \frac{1}{52}\), is equal to the product of their individual probabilities, \(P(E) = \frac{1}{13}\) and \(P(F) = \frac{1}{4}\). This means that drawing an Ace has no effect on the probability of drawing a Diamond and vice versa, as selecting a card of a specific rank does not change the chance of selecting a specific suit.

Step 1: Find the probability of event E

Event E occurs when we draw an Ace from the deck of 52 playing cards. There are 4 Aces in the deck (one for each suit). Thus, the probability of event E can be expressed as: \(P(E) = \frac{\text{number of Aces}}{\text{total number of cards}} = \frac{4}{52} = \frac{1}{13}\)

Step 2: Find the probability of event F

Event F occurs when we draw a Diamond from the deck of 52 playing cards. There are 13 Diamonds in the deck. Thus, the probability of event F can be expressed as: \(P(F) = \frac{\text{number of Diamonds}}{\text{total number of cards}} = \frac{13}{52} = \frac{1}{4}\)

Step 3: Find the probability of both events occurring together and check for independence

Both events E and F occur together when we draw an Ace of Diamonds. There is one Ace of Diamonds in the deck. Thus, the probability of both events occurring together can be expressed as: \(P(E \cap F) = \frac{\text{number of Ace of Diamonds}}{\text{total number of cards}} = \frac{1}{52}\) Now, let's check if events E and F are independent. If they are independent, then the probability of both events occurring together, \(P(E \cap F)\), should be equal to the product of their individual probabilities \(P(E) \times P(F)\). \(P(E \cap F) = P(E) \times P(F)\) By substituting the values calculated earlier, we get: \(\frac{1}{52} = \frac{1}{13} \times \frac{1}{4}\) \(\frac{1}{52} = \frac{1}{52}\) Since both sides of the equation are equal, it is concluded that events E and F are independent. The intuitive explanation for this is that drawing an Ace has no effect on the probability of drawing a Diamond and vice versa. Selecting an Ace card does not change the chance of selecting a specific suit like Diamonds, and selecting a Diamond card does not change the chance of selecting a specific rank, such as an Ace.

Key Concepts

Independent Events

In probability, events are considered independent if the occurrence of one event does not affect the probability of the other event occurring. This means that the outcome of one event has no influence on the other. To determine if events are independent, we use the mathematical condition:

• Two events, say event \(E\) and event \(F\), are independent if and only if \(P(E \cap F) = P(E) \times P(F)\).

This equation states that the probability of both events occurring together is equal to the product of their individual probabilities.
An example of independent events is flipping a coin and rolling a die. The result of the coin flip, heads or tails, does not affect the outcome of the die roll, such as rolling a 4. These two events happening simultaneously still follow the same probability multiplication rule: the probability of getting heads and rolling a 4 is the product of the probability of getting heads and the probability of rolling a 4.

Probability of Intersection

The probability of intersection refers to the chance of two events occurring at the same time. Mathematically, it is denoted by \(P(E \cap F)\), representing the probability that both events \(E\) and \(F\) happen concurrently.
For instance, if you draw a card from a deck, you might want to know the probability of drawing an Ace and a Diamond at the same time. In this case, the intersection event is drawing the "Ace of Diamonds."

• The probability \(P(E \cap F)\) can be calculated directly by counting how many outcomes satisfy both conditions simultaneously and dividing by the total number of possible outcomes.
• From the example, there is only one Ace of Diamonds in a deck, so \(P(E \cap F) = \frac{1}{52}\).

Understanding intersection helps in complex probability scenarios, such as when dealing with compounded events like drawing multiple cards or considering various conditions. Filing a probability for intersection is crucial for decision-making and predicting scenarios accurately.

Probability of a Single Event

The probability of a single event is the chance that a specific event will happen. It is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. This fundamental concept allows one to estimate how likely it is for a specific event to occur, based on the given conditions.
For example, consider the event of drawing an Ace from a standard deck of 52 cards. There are 4 Aces in the deck, so the probability is calculated as:

• \(P(E) = \frac{4}{52} = \frac{1}{13}\)

The same process applies to any single event, such as rolling a die and getting a particular number, like a 3 against the 6 total sides of the dice.

• The probability would then be \(P(A) = \frac{1}{6}\).

Calculating single event probabilities is foundational for understanding more complex probabilistic events and scenarios.

Deck of Cards Probability

A standard deck of cards consists of 52 cards, with 4 suits: Hearts, Diamonds, Clubs, and Spades. Each suit contains 13 cards including numbers from 2 to 10, and the face cards: Jack, Queen, King, and Ace. Understanding deck composition is essential for computing probabilities related to card games and other scenarios involving cards.
When calculating probabilities with a deck, one must consider:

• The total number of cards, which is usually 52 without jokers.
• The number of favorable outcomes, such as the number of Aces or the number of cards within a specific suit.
• The individual probabilities, for single cards or types of cards (e.g., face cards or suits).

Using this, probabilities can be derived, such as:

• Probability of picking a Diamond: \(P( ext{Diamond}) = \frac{13}{52} = \frac{1}{4}\)
• Probability of drawing an Ace: \(P( ext{Ace}) = \frac{4}{52} = \frac{1}{13}\)
• Probability of drawing the Ace of Diamonds: \(P( ext{Ace of Diamonds}) = \frac{1}{52}\)

Deck of cards probability offers practical examples to understand these concepts, serving as a great tool for visualizing independence, intersection, and single events.