Problem 37 - Algebra and Trigonometry with Analytic Geometry

Question

Card and die experiment Each suit in a deck is made up of an ace (A), nine numbered cards \((2,3, \ldots, 10)\), and three face cards (J, Q, K). An experiment consists of drawing a single card from a deck followed by rolling a single die. (a) Describe the sample space \(S\) of the experiment, and find \(n(S)\). (b) Let \(E_{1}\) be the event consisting of the outcomes in which a numbered card is drawn and the number of dots on the die is the same as the number on the card. Find \(n\left(E_{1}\right), n\left(E_{1}^{\prime}\right)\), and \(P\left(E_{1}\right)\). (c) Let \(E_{2}\) be the event in which the card drawn is a face card, and let \(E_{3}\) be the event in which the number of dots on the die is even. Are \(E_{2}\) and \(E_{3}\) mutually exclusive? Are they independent? Find \(P\left(E_{2}\right), P\left(E_{3}\right)\), \(P\left(E_{2} \cap E_{3}\right)\), and \(P\left(E_{2} \cup E_{3}\right)\). (d) Are \(E_{1}\) and \(E_{2}\) mutually exclusive? Are they independent? Find \(P\left(E_{1} \cap E_{2}\right)\) and \(P\left(E_{1} \cup E_{2}\right)\).

Step-by-Step Solution

Answer

Sample space has 312 outcomes, with \( n(E_1) = 9 \). \( E_2 \) and \( E_3 \) are independent; \( E_1 \) and \( E_2 \) are mutually exclusive.

Step 1: Describe the Sample Space

A standard deck of cards has four suits. Each suit has 13 cards: Ace, 2-10, Jack, Queen, King, for a total of 52 cards. A die has 6 faces numbered from 1 to 6. The sample space \( S \) consists of 52 types of drawn cards combined with 6 die results, totaling: \( 52 \times 6 = 312 \). Thus, \( S = \{(\text{card, die result})\} \).

Step 2: Calculate n(S)

The number of possible outcomes in the sample space \( n(S) = 52 \times 6 = 312 \).

Step 3: Calculate n(E1)

\( E_1 \) occurs when the card is a numbered card (2 to 10) and the die shows the same number. There are 9 possible numbered cards (2 to 10), and for each corresponding die roll, only 1 die result matches the card number. So, \( n(E_1) = 9 \).

Step 4: Calculate n(E1')

\( E_1' \) includes all outcomes not in \( E_1 \), so \( n(E_1') = n(S) - n(E_1) = 312 - 9 = 303 \).

Step 5: Calculate P(E1)

The probability of \( E_1 \) is given by: \( P(E_1) = \frac{n(E_1)}{n(S)} = \frac{9}{312} = \frac{3}{104} \).

Step 6: Analyze Events E2 and E3

\( E_2 \) (drawing a face card) means choosing J, Q, or K, for 3 face cards per suit, totaling 12 face cards. Probability: \( P(E_2) = \frac{12}{52} = \frac{3}{13} \). \( E_3 \) (even die number) includes 2, 4, 6, so: \( P(E_3) = \frac{3}{6} = \frac{1}{2} \). \( E_2 \) and \( E_3 \) are not mutually exclusive; they can occur simultaneously. They are independent events since the outcome of one does not affect the other.

Step 7: Calculate P(E2 ∩ E3)

\( P(E_2 \cap E_3) = P(E_2) \times P(E_3) = \frac{3}{13} \times \frac{1}{2} = \frac{3}{26} \), because \( E_2 \) and \( E_3 \) are independent.

Step 8: Calculate P(E2 ∪ E3)

Using the formula \( P(E_2 \cup E_3) = P(E_2) + P(E_3) - P(E_2 \cap E_3) \), we find: \( \frac{3}{13} + \frac{1}{2} - \frac{3}{26} = \frac{12}{52} + \frac{26}{52} - \frac{6}{52} = \frac{32}{52} = \frac{8}{13} \).

Step 9: Relationship Between E1 and E2

\( E_1 \) and \( E_2 \) are mutually exclusive because \( E_1 \) requires a numbered card and \( E_2 \) requires a face card. They can't happen simultaneously, so \( P(E_1 \cap E_2) = 0 \).

Step 10: Calculate P(E1 ∪ E2)

Since \( E_1 \) and \( E_2 \) are mutually exclusive, \( P(E_1 \cup E_2) = P(E_1) + P(E_2) = \frac{3}{104} + \frac{3}{13} = \frac{30}{104} = \frac{15}{52} \).

Key Concepts

Sample Space

In probability, the concept of a sample space is fundamental. It refers to the set of all possible outcomes that can occur in an experiment. Understanding this is crucial because it forms the basis for calculating probabilities. In a card and die experiment, the sample space is formed by combining the outcomes of two separate actions: drawing a card and rolling a die.

• A standard deck contains 52 cards, each with four suits of 13 cards.
• A die has six faces, numbered from 1 to 6.

When these two actions are combined, we create a set of paired outcomes. Each card can result in one of six die numbers, leading to a total of 312 unique outcomes. Thus, for this experiment, the sample space is often described as all possible pairs \((\text{card, die result})\), making the calculation straightforward: \(n(S) = 52 \times 6 = 312\).

Mutually Exclusive Events

When dealing with probability, it's important to understand mutually exclusive events. These are events that cannot occur at the same time. In the context of this card and die experiment, knowing whether events are mutually exclusive helps manage expectations about simultaneous outcomes.

• For example, consider the events \(E_1\) and \(E_2\).
• \(E_1\) is drawing a numbered card where the die matches the card number.
• \(E_2\) is drawing a face card.

Since a card can either be numbered (2 to 10) or a face card (J, Q, K), but not both, \(E_1\) and \(E_2\) are mutually exclusive. This means \(P(E_1 \cap E_2) = 0\), and the probability of either event occurring is simply the sum of their probabilities: \(P(E_1 \cup E_2) = P(E_1) + P(E_2)\).

Independent Events

Independent events are those whose outcomes do not affect each other. Understanding whether events are independent is essential for accurately calculating probabilities. In the card and die experiment, the concept of independent events comes into play when considering the outcomes of drawing a card and rolling a die.

• Event \(E_2\) involves drawing a face card.
• Event \(E_3\) involves rolling an even number on the die (2, 4, or 6).

These two events are independent because the outcome of drawing a card does not influence the roll of the die, and vice versa. Thus, the probability of both events occurring together, \(P(E_2 \cap E_3)\), can be calculated by multiplying their individual probabilities: \(P(E_2) \times P(E_3)\). This property simplifies the calculation of their intersection and union probabilities.

Card and Die Experiment

Probabilistic experiments involving both a card and a die are intriguing because they blend two different random processes. The card and die experiment in this context is designed to explore the probabilities of combined outcomes:

• Drawing a card from a standard 52-card deck.
• Rolling a six-sided die.

Through this setup, we can analyze various events such as matching numbered cards and die outcomes, or drawing specific types of cards like face cards. This experiment showcases not only basic probability rules, like dealing with sample spaces, but also more complex concepts such as identifying whether events are mutually exclusive or independent. Understanding the outcomes of this experiment allows for a practical application of probability theories, helping learners visualize how these mathematical concepts translate into real-world scenarios.