/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 A fair die is rolled 10 times. C... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 6

A fair die is rolled 10 times. Calculate the expected sum of the 10 rolls.

Short Answer

Expert verified
The expected sum of 10 rolls of a fair die is 35.

Step by step solution

01

Determine the expected value of a single roll

The expected value of a single roll is the average value of all possible outcomes. In the case of a fair six-sided die, the possible outcomes are the numbers 1 through 6, and the probability of each outcome is 1/6. The expected value can be calculated using the following formula: Expected Value = 危 (value 脳 probability) For a fair die, the expected value is: Expected Value = \(1(\frac{1}{6}) + 2(\frac{1}{6}) + 3(\frac{1}{6}) + 4(\frac{1}{6}) + 5(\frac{1}{6}) + 6(\frac{1}{6})\)
02

Calculate the expected value of a single roll

Using the above formula, we can now calculate the expected value of a single roll: Expected Value = \(1(\frac{1}{6}) + 2(\frac{1}{6}) + 3(\frac{1}{6}) + 4(\frac{1}{6}) + 5(\frac{1}{6}) + 6(\frac{1}{6})\) Expected Value = \(\frac{1}{6} + \frac{2}{6} + \frac{3}{6} + \frac{4}{6} + \frac{5}{6} + \frac{6}{6}\) Expected Value = \(\frac{21}{6}\) Expected Value = 3.5
03

Calculate the expected sum of 10 rolls

Now that we know the expected value of a single roll (3.5), we can multiply it by the number of rolls (10) to find the expected sum of the 10 rolls: Expected Sum = (Expected Value of a Single Roll) 脳 (Number of Rolls) Expected Sum = 3.5 脳 10 = 35 So, the expected sum of 10 rolls of a fair die is 35.

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Most popular questions from this chapter

A prisoner is trapped in a cell containing 3 doors. The first door leads to a tunnel that returns him to his cell after 2 days' travel. The second leads to a tunnel that returns him to his cell after 4 days' travel. The third door leads to freedom after 1 day of travel. If it is assumed that the prisoner will always select doors \(1,2,\) and 3 with respective probabilities \(.5,3,\) and \(.2,\) what is the expected number of days until the prisoner reaches freedom?

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