Problem 49
Question
A single die is rolled twice. Find the probability of rolling a 2 the first time and a 3 the second time.
Step-by-Step Solution
Verified Answer
The probability of rolling a 2 first and a 3 second is \( \frac{1}{36} \).
1Step 1: Determine the Probability of Rolling a 2
The die has six faces, of which one face is number 2. Therefore, the probability of rolling a 2 is \( \frac{1}{6} \).
2Step 2: Determine the Probability of Rolling a 3
Similarly, one out of six faces of the die is number 3. So, the probability of rolling a 3 is also \( \frac{1}{6} \).
3Step 3: Determine the Combined Probability
The two events are independent. Thus, to determine the combined probability of both events happening, we simply multiply the individual probabilities together. Therefore, the required probability is \( \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \).
Key Concepts
Probability TheoryIndependent Events in ProbabilityProbability Calculation
Probability Theory
Probability theory sits at the very heart of statistics and data analysis, providing a mathematical framework for measuring the likelihood of events. At its simplest, probability represents how likely it is that something will happen, usually expressed as a number between 0 (impossible) and 1 (certain). This theory helps us structure uncertainty and make predictions about the outcomes of various random events.
When we roll a die, for example, the theory guides us to understand that each outcome (1, 2, 3, 4, 5, or 6) has an equally likely chance of occurring, assuming the die is fair. We often use fractions or percentages to articulate this likelihood. For a fair six-sided die, the probability of any specific number rolling out is always \( \frac{1}{6} \), or roughly 16.67%.
When we roll a die, for example, the theory guides us to understand that each outcome (1, 2, 3, 4, 5, or 6) has an equally likely chance of occurring, assuming the die is fair. We often use fractions or percentages to articulate this likelihood. For a fair six-sided die, the probability of any specific number rolling out is always \( \frac{1}{6} \), or roughly 16.67%.
Independent Events in Probability
Independent events are foundational concepts in probability theory, where the outcome of one event does not affect the outcome of another. For example, when flipping a coin, the result of one flip does not influence the result of the next flip. This independence means that we can consider each event separately when calculating probabilities.
In the context of our exercise with rolling dice, the outcome of the first roll does not influence the result of the second roll. Therefore, each roll is an independent event. Understanding this allows for accurate probability calculations for sequences of events, as long as the events do not affect one another.
In the context of our exercise with rolling dice, the outcome of the first roll does not influence the result of the second roll. Therefore, each roll is an independent event. Understanding this allows for accurate probability calculations for sequences of events, as long as the events do not affect one another.
Probability Calculation
Probability calculation is crucial when predicting or understanding the chance of a specific event occurring. The rule for independent events is particularly straightforward – we multiply the probabilities of each independent event to find the likelihood of them all occurring in sequence.
For our two-roll dice scenario, calculating the probability of two specific outcomes involves multiplying the probability of rolling a 2 on the first roll by the probability of rolling a 3 on the second roll. Each has a probability of \( \frac{1}{6} \), giving us \( \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \), which equals approximately 2.78%. It's important for students to grasp that whilst each roll is simple to calculate, combining them requires this multiplication rule which only holds true for independent events.
For our two-roll dice scenario, calculating the probability of two specific outcomes involves multiplying the probability of rolling a 2 on the first roll by the probability of rolling a 3 on the second roll. Each has a probability of \( \frac{1}{6} \), giving us \( \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \), which equals approximately 2.78%. It's important for students to grasp that whilst each roll is simple to calculate, combining them requires this multiplication rule which only holds true for independent events.
Other exercises in this chapter
Problem 48
Write out the first three terms and the last term. Then use the formula for the sum of the first n terms of an arithmetic sequence to find the indicated sum. $$
View solution Problem 48
Graph the hyperbola whose equation is $$ 25 x^{2}-16 y^{2}-100 x-96 y-444=0 $$ Where are the foci located? What are the equations of the asymptotes?
View solution Problem 49
Express each repeating decimal as a fraction in lowest terms. $$ 0 . \overline{257} $$
View solution Problem 49
Use the Binomial Theorem to expand each expression and write the result in simplified form. $$ \left(x^{3}+x^{-2}\right)^{4} $$
View solution