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What is the probability of an odd sum when you roll three dice?

Short Answer

Expert verified
The probability of an odd sum when rolling three dice is \(\frac{3}{4}\).

Step by step solution

01

Determine Possible Outcomes

When rolling three six-sided dice, each die has 6 faces. Therefore, the total number of possible outcomes when rolling three dice is \(6 \times 6 \times 6 = 216\).
02

Understand Conditions for an Odd Sum

An odd sum occurs when the total number from the dice results in an odd number. The sum of the numbers on three dice is odd if two numbers are even and one is odd, or if two numbers are odd and one is even.
03

Calculate Probability for Two Evens and One Odd

For a die, half the numbers (1, 3, 5) are odd, and half (2, 4, 6) are even. Suppose the first two dice are even and the third is odd. The probability is \((\frac{1}{2})^2 \times \frac{1}{2} = \frac{1}{8}\). Since there are 3 different dice that can be odd, multiply by 3: \(3 \times \frac{1}{8} = \frac{3}{8}\).
04

Calculate Probability for Two Odds and One Even

Similarly, for two dice being odd and one even, the probability for any specific combination (two odds, one even) is \((\frac{1}{2})^2 \times \frac{1}{2} = \frac{1}{8}\). Since any one of the three dice can be even, multiply by 3: \(3 \times \frac{1}{8} = \frac{3}{8}\).
05

Combine Probabilities

Add the probabilities from steps 3 and 4 to get the total probability of an odd sum: \(\frac{3}{8} + \frac{3}{8} = \frac{6}{8} = \frac{3}{4}\).
06

Final Calculation and Result

Thus, the probability of rolling an odd sum with three dice is \(\frac{3}{4}\), which means there are 162 favorable outcomes out of 216 possible outcomes.

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Key Concepts

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

Dice Probability
When dealing with dice, understanding the probability of certain outcomes is key. Each standard die has six faces, numbered from 1 to 6. When a die is rolled, each of these faces has an equal chance of landing face up, which gives each outcome a probability of \( \frac{1}{6} \).

Now, when we roll multiple dice, such as in our task with three dice, the possible outcomes multiply. For each die, two dice can roll one of six options too, resulting in \( 6 \times 6 \times 6 = 216 \) total possible combinations. This forms the foundational step in calculating more complex probabilities, such as sums or specific patterns like odds or evens.

Understanding these basics of probability helps us calculate more nuanced problems with dice, where order and specific sequences may play a role in determining the outcome of the roll.
Combinatorics in Probability
Combinatorics is a branch of mathematics that deals with counting, and it is crucial in probability calculations. In probability, we use combinatorics to determine the number of favorable outcomes versus the total number of possible outcomes. This is essential for tasks like determining the probability of rolling an odd sum with dice.

When solving such problems, we often look at combinations of events. For example, with three dice, combinations consist of different states each die can occupy. For an odd sum, we need combinations in which the sum of the numbers is odd. This occurs if there are either two even numbers and one odd, or two odd numbers and one even. Each of these combinations can happen in multiple ways, and this is where counting these possibilities plays a crucial role.

By understanding combinatorics, we can efficiently calculate how many of possible combinations fulfill a specific condition, like having an odd sum, something that is pivotal in many probabilistic problems.
Probability of Sum
In probability, calculating the likelihood of achieving a specific sum, such as an odd or even sum with dice, develops an additional layer of complexity. To find the probability of an odd sum when rolling three dice, we analyze the sum's parity (odd or even nature).

An odd sum can be reached under two conditions: having two numbers as even and one as odd, or two numbers as odd and one as even.
  • Two evens and one odd: Calculate probability for each combination and find it's \( \frac{1}{8}\). Since any die can be the odd one, we multiply by three, giving us \( \frac{3}{8} \).
  • Two odds and one even: The probability follows the same pattern, leading to an additional \( \frac{3}{8} \) probability.
Adding these together provides the total probability of an odd sum: \[ \frac{3}{8} + \frac{3}{8} = \frac{6}{8} = \frac{3}{4} \]. Thus, there is a 75% chance of rolling an odd sum with three dice. This thorough understanding breaks down the task into manageable components, yielding a comprehensive result.

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Most popular questions from this chapter

Suppose a student who knows \(60 \%\) of the material covered in a chapter of a textbook is going to take a five-question objective (each answer is either right or wrong, not multiple choice or true-false) quiz. Let \(X\) be the random variable that gives the number of questions the student answers correctly for each quiz in the sample space of all quizzes the instructor could construct. What is the expected value of the random variable \(X-3 ?\) What is the expected value of \((X-3)^{2}\) ? What is the variance of \(X\) ?

A dime and a 50-cent piece are in a cup. You withdraw one coin. What is the expected amount of money you withdraw? What is the variance? You then draw a second coin, without replacing the first. What is the expected amount of money you withdraw? What is the variance? Suppose instead that you consider withdrawing two coins from the cup together. What is the expected amount of money you withdraw, and what is the variance? What does this example show about whether the variance of a sum of random variables is the sum of their variances?

Give an example of two random variables \(X\) and \(Y\) such that \(E(X Y) \neq E(X) E(Y)\). Here \(X Y\) is the random variable with \((X Y)(s)=X(s) Y(s)\)

Two nickels, two dimes, and two quarters are in a cup. You draw three coins, one after the other, without replacement. What is the expected amount of money you draw on the first draw? On the second draw? What is the expected value of the total amount of money you draw? Does this expected value change if you draw the three coins all at once?

In an independent trials process consisting of six trials with probability p of success, what is the probability that the first three trials are successes and the last three are failures? The probability that the last three trials are successes and the first three are failures? The probability that Trials 1, 3, and 5 are successes and Trials 2, 4, and 6 are failures? What is the probability of three successes and three failures?

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