Problem 42 - Algebra 2

Question

Determine whether the events are mutually exclusive or inclusive. Then find the probability. A card is drawn from a standard deck of cards. What is the probability that it is a 5 or a spade?

Step-by-Step Solution

Answer

The events are inclusive, with a probability of \(\frac{4}{13}\).

Step 1: Define the Events

There are two events: Event A is drawing a 5, and Event B is drawing a spade. We need to determine if these events are mutually exclusive or inclusive. Mutually exclusive events cannot happen at the same time, while inclusive events can.

Step 2: Determine if the Events are Mutually Exclusive or Inclusive

A standard deck has 52 cards, consisting of four suits. Since one of the 5s is a spade, the events are inclusive because drawing a 5 and drawing a spade can happen simultaneously.

Step 3: Calculate the Probability of Each Event

Find the probability of drawing a 5 (Event A): There are 4 fives in a deck (5 of spades, 5 of hearts, 5 of clubs, 5 of diamonds). Therefore, \(P(A) = \frac{4}{52}\). Find the probability of drawing a spade (Event B): There are 13 spades in a deck. Therefore, \(P(B) = \frac{13}{52}\).

Step 4: Calculate the Probability of Intersection

Since the events are inclusive, we need the probability of drawing a card that is both a 5 and a spade (5 of spades). Thus, \(P(A \cap B) = \frac{1}{52}\).

Step 5: Use the Inclusion-Exclusion Principle

The formula for calculating the probability of either event A or event B occurring is \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). Substitute the previously calculated probabilities: \(P(A \cup B) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52}\).

Step 6: Simplify the Probability

Evaluate the expression from Step 5: \(P(A \cup B) = \frac{4+13-1}{52} = \frac{16}{52}\). Simplify the fraction: \(P(A \cup B) = \frac{4}{13}\).

Key Concepts

Mutually exclusive events

Certain events cannot happen simultaneously. This characteristic defines mutually exclusive events. For example, if you roll a single six-sided die, the result cannot simultaneously be a 2 and a 3. When dealing with probability, understanding whether events are mutually exclusive is essential because it directly impacts how probabilities are calculated.

When two events are mutually exclusive, the probability of either occurring is simply the sum of their individual probabilities. Mathematically, this is represented as:

• If events A and B are mutually exclusive, then the probability of either A or B happening is \(P(A \cup B) = P(A) + P(B)\).

In our exercise, the events involve drawing a card from a deck: drawing a 5 and drawing a spade. Because the 5 of spades exists, these events are not mutually exclusive. This means both events can occur at the same time, leading us to explore inclusive events.

Inclusive events

Inclusive events can happen at the same time. In our card example, drawing a 5 and drawing a spade are inclusive events, since you can draw the 5 of spades. The possibility of simultaneous occurrence changes how we calculate probabilities.

When dealing with inclusive events, it's crucial to account for the overlap, or the situation where both events happen. For instance, in our exercise:

• Event A is drawing a 5, and Event B is drawing a spade.
• There is one card, the 5 of spades, which accounts for this overlap.

To find the actual probability when events can overlap, we turn to a helpful principle called the Inclusion-Exclusion Principle, which corrects our calculation by considering this shared event. Without this step, we might overestimate the likelihood of these events occurring.

Inclusion-Exclusion Principle

The Inclusion-Exclusion Principle is a key strategy in probability, especially when dealing with inclusive events. This principle allows us to account for the overlap between events, ensuring that we do not double-count any scenario.

For our card problem, to find the probability of drawing either a 5 or a spade, consider:

• The probability of drawing a 5: \(P(A) = \frac{4}{52}\) because there are 4 fives in the deck.
• The probability of drawing a spade: \(P(B) = \frac{13}{52}\) because there are 13 spades.
• The overlap, 5 of spades, already counted in both previous probabilities: \(P(A \cap B) = \frac{1}{52}\).

The formula to find the probability of either happening is:\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]By applying this, you ensure that the overlapping scenario is not counted twice.

In the solution, we find:\[P(A \cup B) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}\]This simplifying step ensures an accurate probability calculation, adjusting for inclusions.