Question

For the experiments, list the simple events in the sample space, assign probabilities to the simple events, and find the required probabilities. A fair die is tossed twice. What is the probability that the first die is a 6 and the second die is greater than \(2 ?\)

Short answer

Answer: The probability is $\frac{1}{9}$.

Step-by-step solution

List the simple events in the sample space

Since we toss a fair die twice, there are 6 possible outcomes for both first and second dice. As a result, we will have a total of \((6 × 6) = 36\) possible outcomes in our sample space. We can represent these outcomes as ordered pairs: \(S = \{(1,1), (1,2), ... ,(6,6) \}\)

Assign probabilities to the simple events

Since the die is fair, each outcome has the same probability of \(\frac{1}{6}\) for both dice. Therefore, the probability of each pair of outcomes (i, j) is \(P((i, j)) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\) Thus, each ordered pair in the sample space has a probability of \(\frac{1}{36}\).

Identify the targeted event

We are looking for the probability that the first die results in a 6 and the second die results in a number greater than 2. We can represent this event as a subset of the sample space: \(A = \{(6,3), (6,4), (6,5), (6,6) \}\) This event A contains four simple events.

Calculate the required probability

To find the probability of the event A, we sum the probabilities of all simple events in A: \(P(A) = P((6,3)) + P((6,4)) + P((6,5)) + P((6,6))\) As we know, each ordered pair has a probability of \(\frac{1}{36}\), so we get \(P(A) = \frac{1}{36} + \frac{1}{36} + \frac{1}{36} + \frac{1}{36} = \frac{4}{36}\) Finally, we simplify the fraction: \(P(A) = \frac{4}{36} = \frac{1}{9}\) So the probability that the first die is a 6 and the second die is greater than 2 is \(\frac{1}{9}\).

Key concepts

Sample Space

When we talk about probability, one of the fundamental concepts is the "sample space." This is essentially the set of all possible outcomes of a probabilistic experiment. In our exercise, we are tossing a fair die twice, and each toss of the die can land on any of its six faces. Hence, for two tosses, the sample space becomes all possible pairs of these outcomes.

Each outcome is represented as an "ordered pair," where the first number refers to the result of the first die, and the second number refers to that of the second die. Thus, the sample space can be represented as:

• (1,1), (1,2), ..., (1,6),
• (2,1), (2,2), ..., (2,6),
• ⋮
• (6,1), (6,2), ..., (6,6)

This gives us a total of 36 possible outcomes, representing every conceivable scenario when you toss a fair die twice.

Simple Events

A "simple event" is one of the individual possible results within a sample space. Each outcome in the sample space represents a simple event. For example, getting an outcome of (6,3) is a single simple event.

In probability, each simple event is typically assigned equal chances of occurring, provided they are independent and unbiased. In our scenario, with each die toss being distinct and fair, each simple event — like obtaining a result of (1,1) or (6,6) — has the same likelihood of happening.

Understanding simple events is crucial because they are the building blocks of more complex probability events that combine multiple simple events.

Fair Die

The concept of a "fair die" is central to many probability problems. A fair die means that each of the six faces has an equal chance of landing face up after a roll. The probability of any single face landing up is \( \frac{1}{6} \).

When working with a fair die over multiple rolls, the fairness property ensures that the outcomes remain independent from each other. This is why when we roll the die twice, the joint probability of both outcomes being specific numbers is computed as the product of their individual probabilities, given by \( \left( \frac{1}{6} \right) \times \left( \frac{1}{6} \right) = \frac{1}{36} \).

A fair die removes any favorability or bias, lending credibility to our calculations and ensuring our probabilities are accurate and reliable.

Probability Calculation

Probability calculation involves determining the likelihood of a particular event occurring within the defined sample space. Once the sample space and simple events are defined, calculating probability becomes straightforward.

For our specific problem, we want to determine the probability that the first die is a 6 and the second die results in a number greater than 2. The related simple events here are (6,3), (6,4), (6,5), and (6,6).

• We start by noting that for each simple event in our list, the probability is \( \frac{1}{36} \) (since all are equal).
• Next, we sum the probabilities of these simple events because they all represent different ways the desired event can occur:

\[P((6,3)) + P((6,4)) + P((6,5)) + P((6,6)) = \frac{4}{36} = \frac{1}{9}.
\]Thus, the probability that this particular event occurs with a fair die being tossed twice is \( \frac{1}{9} \), signifying a fair chance in probabilistic terms.