Question

A coin is tossed, a die is rolled, and a card is drawn at random from a deck. Assume that the toss, roll, and draw are fair. (a) Describe this experiment as a cross product sample space. (b) With the aid of a tree diagram, define a probability density on the cross product. (c) Verify by direct computation that the probability density found in part (b) is legitimate. (d) Does it matter in what order the coin, the die, and the card are considered?

Short answer

The experiment's sample space has 624 outcomes with equal probability \( \frac{1}{624} \). The event order does not affect the sample space or probabilities.

Step-by-step solution

Define Sample Space Components

1. **Coin toss**: The sample space is \( S_1 = \{H, T\} \), where \( H \) is heads and \( T \) is tails.2. **Die roll**: The sample space is \( S_2 = \{1, 2, 3, 4, 5, 6\} \).3. **Card draw**: A standard deck has 52 cards, so the sample space is \( S_3 = \{1, 2, ..., 52\} \).

Formulate Cross Product Sample Space

The complete sample space is the cross product of the three sample spaces: \[ S = S_1 \times S_2 \times S_3 = \{ (H, 1, 1), ..., (T, 6, 52) \} \]This sample space contains all possible outcomes of the experiment. The total number of outcomes is \( 2 \times 6 \times 52 = 624 \).

Construct Tree Diagram

A tree diagram for this experiment would branch as follows: 1. Start with the coin toss, branching into two outcomes (H, T). 2. From each coin outcome, branch into six outcomes for the die roll (1 through 6). 3. From each die roll outcome, branch into 52 outcomes for each card (1 through 52). Each complete path through the tree represents an outcome from the cross product sample space.

Define Probability Density

Since each event is fair: 1. The probability of each coin outcome is \( P(H) = P(T) = \frac{1}{2} \).2. The probability of each die face is \( P(1) = P(2) = ... = P(6) = \frac{1}{6} \).3. The probability of drawing any specific card is \( \frac{1}{52} \).Thus, the probability of any particular outcome \( (c, d, a) \) is: \[ P(c, d, a) = P(c) \times P(d) \times P(a) = \frac{1}{2} \times \frac{1}{6} \times \frac{1}{52} = \frac{1}{624} \].

Verify Probability Density

To verify, sum the probabilities of all outcomes:\[ \sum_{(c, d, a) \in S} P(c, d, a) = \sum_{i=1}^{624} \frac{1}{624} = 1 \].The total probability is 1, confirming it is a legitimate probability distribution.

Evaluate Order of Events

The order does not matter, as each stage in the sequence (coin toss, die roll, card draw) is independent. The definition of the cross product and the resultant sample space remain unchanged regardless of the sequence of events.

Key concepts

Sample Space

When dealing with probability, a sample space is a fundamental concept. It refers to the set of all possible outcomes of a particular experiment. In our scenario, the experiment consists of a coin toss, a die roll, and a card draw, each with its unique sample space.
- **Coin Toss**: The sample space, denoted as \( S_1 \), is \( \{H, T\} \), where \( H \) stands for heads and \( T \) stands for tails.
- **Die Roll**: Here, the sample space \( S_2 \) is \( \{1, 2, 3, 4, 5, 6\} \), representing each face of a six-sided die.
- **Card Draw**: For drawing a card from a standard deck of 52 cards, the sample space \( S_3 \) is \( \{1, 2, ..., 52\} \), each number representing a unique card.
By taking the cross product of these spaces—\( S = S_1 \times S_2 \times S_3 \)—we obtain a comprehensive sample space that contains all possible outcomes of this combined experiment. There are 624 possible outcomes in total, calculated by multiplying the number of outcomes from each individual sample space: \( 2 \times 6 \times 52 = 624 \).

Probability Distribution

A probability distribution describes how probabilities are assigned to each possible outcome in a sample space. For the experiment in question, each element of the sample space is assigned a probability.
Given that the coin toss, die roll, and card draw are all fair events, we assign probabilities based on the formulas:

• Each side of the coin has a probability of \( \frac{1}{2} \).
• Each face of the die has a probability of \( \frac{1}{6} \).
• Each card in the deck has a probability of \( \frac{1}{52} \).

With these probabilities, the probability of any specific outcome, for instance, obtaining \((H, 3, 10)\), is calculated by multiplying the probabilities of each individual event:\[ P(H, 3, 10) = \frac{1}{2} \times \frac{1}{6} \times \frac{1}{52} = \frac{1}{624} \] This consistent assignment of probability ensures that the sum of all probabilities across the sample space is 1, fulfilling the conditions for a legitimate probability distribution.

Tree Diagram

A tree diagram is a powerful visual tool that helps in organizing and visualizing the outcomes of a complex experiment. It branches out step by step from the start of the experiment to the end. For the experiment involving a coin toss, die roll, and card draw, the tree diagram illustrates potential outcomes:
1. **Start with Coin Toss**: The tree splits into two branches, one for heads (H) and another for tails (T).
2. **Die Roll**: Each outcome from the coin toss (H or T) branches into six more outcomes, corresponding to the numbers 1 to 6 on a die.
3. **Card Draw**: Each die outcome branches into 52 possibilities, one for each card in a standard deck.
Each pathway from the start to the final branching represents an outcome in our sample space. A tree diagram simplifies the calculation process by breaking down the experiment into manageable steps and clearly showing all possible outcomes.

Independent Events

In probability, independent events are those whose outcomes do not influence each other. When two events are independent, the outcome of one event has no effect on the outcome of the other. This key concept is relevant to our scenario.
Each segment of the combined experiment—the coin toss, the die roll, and the card draw—operates independently. The result of tossing the coin doesn’t affect the result of rolling the die, and vice versa, nor does it affect drawing a card from the deck.
According to the multiplication rule for independent events, the overall probability of a combination of these independent events is the product of their individual probabilities. Therefore, it's concluded that:\[ P(H, 3, 10) = P(H) \times P(3) \times P(10) \] Reassuringly, independence in this context means that the order in which the events occur—whether you toss the coin first, roll the die next, or draw the card last—does not alter the overall probability result. This adds a layer of simplification to calculating complex probability scenarios.