Domanda del libro di testo
Express each repeating decimal as a fraction in lowest terms.
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Ch. 8 - Sequences, Induction, and Probability
Blitzer8th EditionCollege AlgebraISBN: 9780136970514
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TOC.tableOfContents
- Ch. P - Fundamental Concepts of Algebra311
- Ch. 1 - Equations and Inequalities398
- Ch. 2 - Functions and Graphs407
- Ch. 3 - Polynomial and Rational Functions306
- Ch. 4 - Exponential and Logarithmic Functions247
- Ch. 5 - Systems of Equations and Inequalities173
- Ch. 6 - Matrices and Determinants143
- Ch. 7 - Conic Sections124
- Ch. 8 - Sequences, Induction, and Probability251
Capitolo 9, Problema 49
In Exercises 49–52, a single die is rolled twice. Find the probability of rolling a 2 the first time and a 3 the second time.
Guida verificata passo dopo passo1
Understand that the problem involves two independent events: rolling a 2 on the first roll and rolling a 3 on the second roll of a single die.
Recall that the probability of rolling any specific number on a fair six-sided die is \(\frac{1}{6}\).
Calculate the probability of rolling a 2 on the first roll, which is \(\frac{1}{6}\).
Calculate the probability of rolling a 3 on the second roll, which is also \(\frac{1}{6}\).
Since the two rolls are independent, multiply the probabilities of each event to find the combined probability: \(\frac{1}{6} \times \frac{1}{6}\).
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Probability of a Single Event
Probability measures the likelihood of an event occurring, calculated as the ratio of favorable outcomes to total possible outcomes. For a fair six-sided die, the probability of rolling any specific number, like a 2, is 1/6.
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5:14
Probability of Mutually Exclusive Events
Independent Events
Two events are independent if the outcome of one does not affect the outcome of the other. Rolling a die twice involves independent events because the result of the first roll does not influence the second.
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5:54Probability of Multiple Independent Events
Multiplication Rule for Independent Events
When two events are independent, the probability of both occurring is the product of their individual probabilities. To find the chance of rolling a 2 first and a 3 second, multiply 1/6 by 1/6, resulting in 1/36.
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5:54Probability of Multiple Independent Events
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