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What is the expected number of times a 6 appears when a fair die is rolled 10 times?

Short Answer

Expert verified
1.67

Step by step solution

01

Understand the Problem

The question asks for the expected number of times a 6 appears when rolling a fair die 10 times. This involves calculating the expected value for a specific outcome in a series of independent trials.
02

Define the Probability

A fair die has 6 faces, numbered 1 through 6. The probability of rolling a 6 on any single roll is \( P(6) = \frac{1}{6} \).
03

Use the Expected Value Formula

For a binomial distribution, the expected value \( E \) can be calculated by multiplying the probability of success \( p \) by the number of trials \( n \). Here, \( n = 10 \) and \( p = \frac{1}{6} \). So, \( E = n \cdot p \).
04

Calculate the Expected Value

Substitute the values into the expected value formula: \( E = 10 \times \frac{1}{6} \). Therefore, \( E = \frac{10}{6} = \frac{5}{3} = 1.67 \).

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Key Concepts

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

expected value
The concept of expected value is essential in probability and statistics. It represents the average outcome that one can anticipate from a random event after many trials. Imagine rolling a fair die multiple times; while you might not always roll a 6, if you roll it enough times, the average number of 6s you get will converge to a predictable value. This value is what we call the expected value.
Calculating the expected value is straightforward. For a binomial distribution, which involves a series of independent trials (like rolling a die multiple times), the expected value \( E \) is found using the formula: \[ E = n \times p \] where \( n \) is the number of trials and \( p \) is the probability of success in a single trial. In our example with the die, the probability of rolling a 6 in one trial is \( \frac{1}{6} \). So, when rolling the die 10 times, the expected value of rolling a 6 is \( 10 \times \frac{1}{6} = 1.67 \).
binomial distribution
The binomial distribution represents the number of successes in a fixed number of independent trials, with each trial having the same probability of success. It is a popular concept in statistics due to its wide applicability in scenarios like flipping coins, rolling dice, or even more complex events such as quality control in manufacturing.
For a scenario to follow a binomial distribution, it must meet these conditions:
  • Each trial is independent.
  • There are exactly two possible outcomes: success or failure.
  • The probability of success \( p \) remains constant for each trial.

When using the binomial distribution, we use parameters such as the number of trials (\ref means_trials\ref), the number of successes (\ref means_successes\ref), and the probability of success (\ref means_success_proba\ref). Understanding these will help you recognize and correctly apply the binomial distribution to different problems. In the case of rolling a fair die, each roll is independent, the outcome of rolling a 6 is considered a success, and the probability remains constant at \( \frac{1}{6} \). Thus, the expected number of 6s in 10 rolls can be calculated using binomial distribution concepts.
probability
Probability is a measure of how likely an event is to occur. It ranges from 0 (impossible event) to 1 (certain event). Understanding probability is vital in everyday decision-making and in fields such as finance, insurance, medicine, and more.
In probability theory, outcomes are essentially the

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Most popular questions from this chapter

In this exercise we will use Bayes’ theorem to solve the Monty Hall puzzle (Example 10 in Section 7.1). Recall that in this puzzle you are asked to select one of three doors to open. There is a large prize behind one of the three doors and the other two doors are losers. After you select a door, Monty Hall opens one of the two doors you did not select that he knows is a losing door, selecting at random if both are losing doors. Monty asks you whether you would like to switch doors. Suppose that the three doors in the puzzle are labeled 1, 2, and 3. Let W be the random variable whose value is the number of the winning door; assume that p(W = k) = 1?3 for k = 1, 2, 3. Let M denote the random variable whose value is the number of the door that Monty opens. Suppose you choose door i. a) What is the probability that you will win the prize if the game ends without Monty asking you whether you want to change doors? b) Find p(M = j ? W = k) for j = 1, 2, 3 and k = 1, 2, 3. c) Use Bayes’ theorem to find p(W = j ? M = k) where i and j and k are distinct values. d) Explain why the answer to part (c) tells you whether you should change doors when Monty gives you the chance to do so.

Suppose that we roll a fair die until a 6 comes up or we have rolled it 10 times. What is the expected number of times we roll the die?

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Use Chebyshev's inequality to find an upper bound on the probability that the number of tails that come up when a biased coin with probability of heads equal to 0.6 is tossed \(n\) times deviates from the mean by more than \(\sqrt{n}\) .

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