Problem 43 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

An experiment consists of selecting a card at random from a well-shuffled 52 -card deck. Let \(E\) denote the event that an ace is drawn and let \(F\) denote the event that a spade is drawn. Show that \(n(E \cup F)=n(E)+n(F)-n(E \cap F)\).

Step-by-Step Solution

Answer

In a well-shuffled 52-card deck, there are 16 favorable outcomes for drawing either an Ace or a Spade, as calculated using the formula \(n(E \cup F) = n(E) + n(F) - n(E \cap F) = 4 + 13 - 1 = 16\).

Step 1: Identify the number of favorable outcomes for drawing an Ace

There are 4 Aces in a standard 52-card deck (one Ace per suit). Thus, \(n(E) = 4\).

Step 2: Identify the number of favorable outcomes for drawing a Spade

There are 13 Spades in a standard 52-card deck (one for each rank). Thus, \(n(F) = 13\).

Step 3: Identify the number of favorable outcomes for drawing an Ace of Spades

There is only 1 Ace of Spades in a standard 52-card deck. Thus, \(n(E \cap F) = 1\).

Step 4: Use the formula to calculate the number of favorable outcomes for drawing either an Ace or a Spade

Plug the values of \(n(E)\), \(n(F)\), and \(n(E \cap F)\) into the formula: \(n(E \cup F) = n(E) + n(F) - n(E \cap F) = 4 + 13 - 1 = 16\) Hence, there are 16 favorable outcomes for drawing either an Ace or a Spade.

Key Concepts

Understanding Favorable Outcomes in Probability

To master probability in card games, one must first understand what favorable outcomes are. In a probability scenario, 'favorable outcomes' are the specific results of an experiment or event that align with a desired condition. For example, if you're interested in drawing an ace from a standard deck of 52 cards, any ace that you draw is considered a favorable outcome.

In the provided exercise, the event E (drawing an ace) has 4 favorable outcomes because there are 4 aces in a deck—one in each suit. Similarly, for event F (drawing a spade), there are 13 favorable outcomes corresponding to the 13 spades. Recognizing the number of favorable outcomes is the first step in solving most probability problems and gives you a foundation on which to apply probability formulas.

Applying Probability Formulas

Calculating probabilities often involves applying specific formulas. One widely used formula is the addition rule, which relates to the probability of the union of two events. The union of events E (drawing an ace) and F (drawing a spade) is denoted as \(E \cup F\), which covers any outcome that is either an ace or a spade—or both.

Understanding the Addition Rule

The basic addition formula in probability is:\[P(E \cup F) = P(E) + P(F) - P(E \cap F)\]This equation factors in the overlap between E and F to avoid double-counting. In our exercise, the calculation showed that \(n(E \cup F) = 16\), indicating 16 favorable outcomes out of 52 when drawing either an ace or a spade, thus providing a clear path to the event's probability. Understanding and correctly applying probability formulas is crucial in determining the likelihood of various outcomes in card games.

Set Theory in Probability

Set theory is inherently related to probability, as it deals with the study of collections of objects, which in probability are events and their outcomes. In the context of card games, these sets are the possible outcomes, and applying set theory principles, like intersections and unions, can solve complex problems.

Intersections and Unions

In our sample problem, we use set theory concepts such as the intersection (\(E \cap F\)) to represent outcomes that both events share—in this case, the single Ace of Spades. The union (\(E \cup F\)) encompasses all unique outcomes from both events (any ace or any spade).

Understanding set theory in probability allows you to calculate the total number of favorable outcomes by properly accounting for shared outcomes between sets. This is paramount when dealing with multiple events and can help avoid common mistakes such as overcounting overlapping outcomes.