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If you were to roll a fair die 1,000 times, about how many sixes do you think you would observe? What is the probability of observing a six when a fair die is rolled?

Short Answer

Expert verified
You are expected to roll a six approximately 167 times out of 1,000 rolls, and the probability of rolling a six in a single roll of a fair die is approximately 0.167 or 1/6.

Step by step solution

01

Calculate Expected Number of Sixes

To calculate the expected number of times a six will be rolled, multiply the total number of trials by the likelihood of a six being rolled in a single trial. The total number of trials is 1,000, and the probability of rolling a six once is 1/6. So, \( 1000 * \frac{1}{6} = 166.67 \). Therefore, a six is expected to come up approximately 167 times.
02

Calculate Probability of Rolling a Six

Since a die is a fair die with six equally likely outcomes, the probability of rolling a six is simply 1/6 or approximately 0.167.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value
When rolling a fair die, the concept of 'expected value' becomes a pivotal tool in understanding what results we can anticipate over the long run. In essence, expected value is the average outcome or result you can expect after running an experiment a large number of times. It’s calculated by multiplying the value of each possible outcome by the probability of that outcome occurring, and then summing these products.

For instance, in the context of our dice-rolling exercise, we want to find out how many times we can expect to get a six if we roll the die 1,000 times. Each roll is independent, and the chance of landing a six is 1 in 6. Mathematically, the expected value (E) equals the total trials (n) times the probability (p) of getting a six, thus: \[ E = n \times p = 1000 \times \frac{1}{6} \.\] Simplifying this yields \[ E \approx 166.67 \.\] So we round this number to the nearest whole number, as we can't roll a fraction of a time. That means you can expect to roll a six approximately 167 times in 1000 rolls.
Fair Die
The term 'fair die' guarantees that all outcomes are equally likely. If you imagine a standard six-sided die, each face numbered 1 through 6, a fair die means each number has an equal chance of landing face up on any given roll. This is the foundation of probability theory where fair means unbiased, unweighted, and not tampered with, ensuring each number has an exact \( \frac{1}{6} \) chance of occurring.

To understand if a die is fair, we can perform a large number of rolls and record the outcomes. If the die is fair, the number of times each of the six numbers comes up should be roughly equal. In the case of the exercise, rolling the die 1,000 times provides a large enough sample size to get close to the theoretical probability. Trusting in the fairness of the die, we use it to further calculate the expected values and probabilities of different outcomes.
Trial Probability
Trial probability is another term for the likelihood of an event occurring in a single instance or trial. In a dice game, each roll is considered a trial, and the probability of an event, such as rolling a six, is determined by the number of favorable outcomes divided by the total number of possible outcomes. On a fair six-sided die, the probability of rolling any given number is \( \frac{1}{6} \) given that there is one favorable outcome and six possible outcomes.

The trial probability remains constant for each roll as long as the die is fair. This concept is crucial in calculating the expected number of times a particular outcome will occur after a series of trials, as well as in understanding that each roll is independent of the others. Regardless of what you rolled before, the trial probability for the next roll is always the same when dealing with a fair die.

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Most popular questions from this chapter

A study of how people are using online services for medical consulting is described in the paper "Internet Based Consultation to Transfer Knowledge for Patients Requiring Specialized Care" (British Medical Journal [2003]: \(696-699)\). Patients using a particular online site could request one or both (or neither) of two services: specialist opinion and assessment of pathology results. The paper reported that \(98.7 \%\) of those using the service requested a specialist opinion, \(35.4 \%\) requested the assessment of pathology results, and \(34.7 \%\) requested both a specialist opinion and assessment of pathology results. Consider the following two events: \(S=\) event that a specialist opinion is requested \(A=\) event that an assessment of pathology results is requested a. What are the values of \(P(S), P(A)\), and \(P(S \cap A)\) ? b. Use the given probability information to set up a "hypothetical 1000 " table with columns corresponding to \(S\) and not \(S\) and rows corresponding to \(A\) and \(\operatorname{not} A .\) c. Use the table to find the following probabilities: i. the probability that a request is for neither a specialist opinion nor assessment of pathology results. ii. the probability that a request is for a specialist opinion or an assessment of pathology results.

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A large cable company reports that \(42 \%\) of its customers subscribe to its Internet service, \(32 \%\) subscribe to its phone service, and \(51 \%\) subscribe to its Internet service or its phone service (or both). a. Use the given probability information to set up a "hypothetical \(1000 "\) table. b. Use the table to find the following: i. the probability that a randomly selected customer subscribes to both the Internet service and the phone service. ii. the probability that a randomly selected customer subscribes to exactly one of the two services.

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