Problem 55 - Algebra and Trigonometry

Question

A die is rolled twice. Let \(E\) and \(F\) be the following events: $$\begin{array}{l}{\text { E. The first roll shows a six. }} \\ {\text { F. The second roll shows a six. }}\end{array}$$ (a) Are the events \(E\) and \(F\) independent? (b) Find the probability of showing a six on both rolls.

Step-by-Step Solution

Answer

(a) Yes, E and F are independent. (b) Probability is \(\frac{1}{36}\).

Step 1: Define Independence of Events

Two events are independent if the occurrence of one event does not affect the occurrence of the other. Mathematically, events \(E\) and \(F\) are independent if \(P(E \cap F) = P(E) \cdot P(F)\).

Step 2: Calculate Probability of Event E

The probability of event \(E\), which is the first roll showing a six, is \(\frac{1}{6}\) because there are 6 possible outcomes and only one favorable outcome.

Step 3: Calculate Probability of Event F

The probability of event \(F\), which is the second roll showing a six, is similarly \(\frac{1}{6}\) for the same reasons as event \(E\).

Step 4: Calculate Probability of Both Events E and F

The probability that both events \(E\) and \(F\) occur, or \(P(E \cap F)\), is the probability of getting a six on both rolls. Since the rolls are independent, we multiply the probabilities: \(P(E \cap F) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}\).

Step 5: Verify Independence of Events

Finally, since \(P(E \cap F) = \frac{1}{36}\) and \(P(E) \cdot P(F) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}\), events \(E\) and \(F\) are independent.

Step 6: Probability of Both Rolls Showing Six

To find the probability of showing a six on both rolls, we calculated \(P(E \cap F) = \frac{1}{36}\).

Key Concepts

Independent Events

In probability theory, understanding whether events are independent is crucial. Two events, such as those from dice rolls, are considered independent if the outcome of one does not affect the probability of the other occurring. This means that knowing the result of one event gives you no information about the other. For example, in the case where a die is rolled twice, the result of the first roll does not influence the result of the second. To confirm independence mathematically, we use the formula:

• \( P(E \cap F) = P(E) \cdot P(F) \)

If this equality holds true, the events are independent. Here, \(P(E \cap F)\) represents the probability of both events occurring together. If this equals the product of their individual probabilities, \(P(E)\) and \(P(F)\), then independence is confirmed.

Probability Calculation

Calculating probabilities can be straightforward when you break it down into simple steps. Each roll of a die is an event with several possible outcomes, and for a fair six-sided die, each face has an equal chance of appearing. When asked to find the probability that a six appears on a single roll, the probability, \(P(E)\) or \(P(F)\), would be:

• \( \frac{1}{6} \)

This calculation is the result of one favorable outcome over six possible outcomes. To find the probability that two independent events, like two rolls, both produce a six, multiply the probabilities of the individual events:

• \( P(E \cap F) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \)

Understanding this multiplication aligns with the concept of probability for independent events. The overall probability is smaller, reflecting the increased difficulty of both independent events occurring simultaneously.

Outcomes of Dice Rolls

The outcomes of dice rolls can be analyzed effectively using probability. With a standard six-sided die, each roll presents six possible outcomes, typically noted as the numbers 1 through 6. Each number has an equal chance, which makes a fair die great for illustrating probability concepts.When rolling a die twice, each roll remains independent, meaning:

• The result of the first roll does not affect the second.

The combined list of potential outcomes for two dice rolls is the Cartesian product of the outcomes of a single die, resulting in 36 total outcomes (since 6 outcomes from the first roll \(\times\) 6 from the second). This expansive set of possibilities helps in calculating joint probabilities like rolling a number or achieving a pair of sixes, where only one of these 36 outcomes (\((6,6)\)) results in both dice showing a six.