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Chapter 13: Problem 24

Find the distribution of the sum of the numbers when a pair of dice is tossed.

Short Answer

Expert verified
The distribution shows probabilities for sums 2-12 from two dice: 2-\(\frac{1}{36}\), 7-\(\frac{6}{36}\), and symmetrical decreases on each side.

Step by step solution

01

Understand the Problem

We are asked to find the distribution of the sum when two dice are rolled. Each die has 6 faces numbered from 1 to 6. We need to calculate all possible sums.
02

Identify Possible Outcomes

Each die has 6 outcomes (1 to 6). Therefore, when two dice are tossed, the number of possible outcomes is \( 6 \times 6 = 36 \).
03

List All Possible Sums

The smallest sum is 2 (1+1) and the largest sum is 12 (6+6). Possible sums are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12.
04

Count Frequency of Each Sum

- Sum of 2: 1 outcome (1+1) - Sum of 3: 2 outcomes (1+2, 2+1) - Sum of 4: 3 outcomes (1+3, 2+2, 3+1) - Sum of 5: 4 outcomes (1+4, 2+3, 3+2, 4+1) - Sum of 6: 5 outcomes (1+5, 2+4, 3+3, 4+2, 5+1) - Sum of 7: 6 outcomes (1+6, 2+5, 3+4, 4+3, 5+2, 6+1) - Sum of 8: 5 outcomes (2+6, 3+5, 4+4, 5+3, 6+2) - Sum of 9: 4 outcomes (3+6, 4+5, 5+4, 6+3) - Sum of 10: 3 outcomes (4+6, 5+5, 6+4) - Sum of 11: 2 outcomes (5+6, 6+5) - Sum of 12: 1 outcome (6+6)
05

Create a Probability Distribution

Calculate the probability for each sum by dividing the number of outcomes for that sum by the total number of possible outcomes (36). For instance, \( P(2) = \frac{1}{36} \), \( P(3) = \frac{2}{36} \), and so on. | Sum | Probability | |-----|-------------|| 2 | \( \frac{1}{36} \) || 3 | \( \frac{2}{36} \) || 4 | \( \frac{3}{36} \) || 5 | \( \frac{4}{36} \) || 6 | \( \frac{5}{36} \) || 7 | \( \frac{6}{36} \) || 8 | \( \frac{5}{36} \) || 9 | \( \frac{4}{36} \) || 10 | \( \frac{3}{36} \) || 11 | \( \frac{2}{36} \) || 12 | \( \frac{1}{36} \) |

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Key Concepts

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

Sum of Dice
When you toss two dice, each die contributes to the overall sum. The smallest possible sum is 2, achieved by rolling a 1 on both dice. The largest sum is 12, when each die shows a 6. The fun part is figuring out what happens in between. This is important because each sum occurs with a different frequency, and understanding this can help in games or probability problems. The sums you can roll are between 2 and 12, making for a total of 11 possible sums. Each of these sums has a different number of ways to be rolled, which leads us into the next fundamental concept: outcomes in probability.
Outcomes in Probability
In probability theory, an outcome is a possible result of a random experiment. For rolling a pair of dice, each die has 6 faces, and each face is a potential outcome. So, the total number of outcomes is obtained by multiplying the number of outcomes of each die:
  • The first die has 6 possible outcomes.
  • The second die also has 6 possible outcomes.
  • Therefore, combined, you have a total of 36 outcomes from rolling two dice: 6 from the first die and 6 from the second die.
These 36 outcomes are the basis for calculating the sums of two dice, as each combination contributes to one of the sums ranging from 2 to 12.
Frequency Distribution
Frequency distribution is a helpful way to understand how often each sum can occur when rolling two dice. For each possible sum between 2 and 12, you can count how many combinations of the dice result in that sum. This gives you a frequency, laying the groundwork for calculating probabilities:
  • Sum of 2: 1 combination
  • Sum of 3: 2 combinations
  • Sum of 4: 3 combinations
  • Sum of 5: 4 combinations
  • Sum of 6: 5 combinations
  • Sum of 7: 6 combinations
  • Sum of 8: 5 combinations
  • Sum of 9: 4 combinations
  • Sum of 10: 3 combinations
  • Sum of 11: 2 combinations
  • Sum of 12: 1 combination
Using this frequency distribution, we can seamlessly transition into creating a probability distribution for the sums.
Probability Theory
Probability theory is the branch of mathematics that deals with the likelihood of different outcomes. With dice, it's all about figuring out how likely it is to roll each sum. To create a probability distribution, you take the frequency of each sum (from the frequency distribution) and divide by the total number of outcomes, which is 36:
  • For a sum of 2: Probability is \( \frac{1}{36} \)
  • For a sum of 3: Probability is \( \frac{2}{36} \)
  • For a sum of 4: Probability is \( \frac{3}{36} \)
  • For a sum of 5: Probability is \( \frac{4}{36} \)
  • Continuing this way up to a sum of 12: Probability is \( \frac{1}{36} \)
This distribution helps you understand not only which sums are possible, but also how likely they are. For example, a sum of 7 is the most likely with a probability of \( \frac{6}{36} \), which makes it the center point in the bell curve of possible sums. That's the power of probability theory to interpret the randomness and likelihood of events.

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Most popular questions from this chapter

The City Engineer's department installs 10000 fluorescent lamp bulbs in street lamp standards. The bulbs have an average life of 7000 operating hours with a standard deviation of 400 hours. Assuming that the life of the bulbs, \(L\), is a normal random variable, what number of bulbs might be expected to have failed after 6000 operating hours? If the engineer wishes to adopt a routine replacement policy which ensures that no more than \(5 \%\) of the bulbs fail before their routine replacement, after how long should the bulbs be replaced?

A fleet car operator has \(n\) cars, each of which has probability \(8 \%\) of being broken down on any particular day. Find the smallest value of \(n\) that gives probability \(90 \%\) that at least forty cars will be available for use on any one day.

During the repair of a large number of car engines it was found that part number 100 was changed in \(36 \%\) and part number 101 in \(42 \%\) of cases, and that both parts were changed in \(30 \%\) of cases. Is the replacement of part 100 connected with that of part \(101 ?\) Find the probability that in repairing an engine for which part 100 has been changed it will also be necessary to replace part 101 .

Suppose that the running distance (in thousands of kilometres) that car owners get from a tyre is a random variable with density function. $$ f_{X}(x)=\left\\{\begin{array}{cc} \frac{1}{30} \mathrm{e}^{-x / 30} & (x>0) \\ 0 & (x \leqslant 0) \end{array}\right. $$ Find (a) the probability that one of these tyres will last at most \(19000 \mathrm{~km}\); (b) the mean and standard deviation of \(X\); (c) the median and interquartile range of \(X\).

Suppose that \(X\) is a continuous random variable with mean \(\mu_{X}\) and variance \(\sigma_{X}^{2}\). By separating the integral in the definition of \(\sigma_{X}^{2}\) into three parts and substituting the respective bounds for \(\left(x-\mu_{X}\right)^{2}\) as follows $$ \left(x-\mu_{X}\right)^{2} \geqslant\left\\{\begin{array}{cl} \left(k \sigma_{x}\right)^{2} & \text { on }\left(-\infty, \mu_{X}-k \sigma_{X}\right) \\ 0 & \text { on }\left(\mu_{x}-k \sigma_{X}, \mu_{X}+k \sigma_{X}\right) \\ \left(k \sigma_{X}\right)^{2} & \text { on }\left(\mu_{X}+k \sigma_{X},+\infty\right) \end{array}\right. $$ where \(k\) is a constant, prove Chebyshev's theorem $$ P\left(\left|x-\mu_{x}\right|>k \sigma_{X}\right) \leqslant k^{-2} $$ Deduce that for every continuous random variable \(X\) the probability is at least \(\frac{8}{9}\) that \(X\) will take a value within three standard deviations of the mean.
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