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Suppose you roll two 12 -sided dice with faces numbered 1 to \(12 .\) a. How many possible number pairs can you roll? b. What is the greatest sum possible from a roll of two 12 -sided dice? c. What sum is most likely? What is the probability of this sum?

Short Answer

Expert verified
a. 144 pairs. b. Greatest sum is 24. c. The most likely sum is 13 with a probability of \(\frac{1}{12}\).

Step by step solution

01

Identify Total Pair Combinations

Each dice has 12 faces, so the total number of possible pairs when rolling two dice is found by multiplying the number of faces of the first dice by the number of faces of the second dice: Number of pairs = 12 * 12 = 144.
02

Determine the Greatest Possible Sum

The greatest possible sum when obtaining the highest numbers on both dice is the sum of the maximum face value of each die: Greatest sum = 12 + 12 = 24.
03

Identify the Most Likely Sum

For two 12-sided dice, the sums range from 2 (1+1) to 24 (12+12). The most likely sum is found at the midpoint of the possible sums, since the distribution of sums forms a symmetrical pattern. The midpoint sum for two dice is calculated as follows:Most likely sum = \[\frac{2 + 24}{2} = 13\].
04

Calculate the Probability of the Most Likely Sum

To find the probability of rolling the most likely sum of 13, count the pairs that result in this sum: - 1 + 12- 2 + 11- 3 + 10- 4 + 9- 5 + 8- 6 + 7- 7 + 6- 8 + 5- 9 + 4- 10 + 3- 11 + 2- 12 + 1There are 12 possible ways to roll a sum of 13. Therefore, the probability is:\[ \text{Probability} = \frac{12}{144} = \frac{1}{12} \text{ or approximately } 0.0833.\]

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Key Concepts

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

Probability Calculation
Probability calculation is the process used to determine how likely an event is to occur. When rolling two 12-sided dice, each die has 12 faces, numbered 1 through 12. Each face is equally likely to land face up. This means each face has a probability of \(\frac{1}{12}\).

When calculating the total number of possible outcomes for two dice, you multiply the number of faces on the first die by the number of faces on the second die:
\[ \text{Number of Pairs} = 12 \times 12 = 144 \] This means there are 144 unique combinations of rolls.

To find the probability of a specific sum, like 13, we count how many ways we can get that sum. For instance, to get 13:
- Roll (1, 12)
- Roll (2, 11)
- Roll (3, 10)
- And so on, up to (12, 1)
There are 12 ways to get this sum. 

So, the probability of rolling a sum of 13 is:
\[ \text{Probability} = \frac{12}{144} = \frac{1}{12} = 0.0833 \] Therefore, the probability is about 8.33%.
Sum of Dice Rolls
The sum of dice rolls is simply the total value when adding the faces of both dice. For two 12-sided dice, the smallest sum occurs when both dice land on 1 and is \(\text{1+1=2}\).
The greatest possible sum is when both dice land on 12, so we get:
\[ \text{Greatest sum} = 12 + 12 = 24 \]

When considering which sum is most likely, we look at the distribution of all possible sums. This distribution forms a symmetrical pattern. The midpoint between the smallest and greatest sum gives us the most likely sum:
\[ \text{Most likely sum} = \frac{2 + 24}{2} = 13 \] This is because the pairs of numbers around this midpoint have the highest number of combinations.

Thus, the sum of 13 is the most probable result when rolling two 12-sided dice.
Dice Combinations
Dice combinations refer to the unique pairs you can get when rolling two dice. For example, a roll of (3, 4) is different from (4, 3), making each pair unique.
When rolling two dice, each die has 12 outcomes. Together, these create a total of:
\[ \text{Number of pairs} = 12 \times 12 = 144 \] combinations.

To understand this concept better, think about listing all numbers from 1 to 12 on the first die and repeating this for the second die. Pairing each number from the first die with each from the second results in 144 different pairs.

When focusing on a specific sum, like 13, remember that different pairs result in the same sum. For instance:
(1, 12), (2, 11), (3, 10) ... up to (12, 1).
Each of these pairs totals to 13, and we've identified 12 such pairs.

So, the dice combinations help us understand how many ways a specific outcome can be reached, which is essential for calculating probabilities.

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Most popular questions from this chapter

Challenge In Investigation \(2,\) you always assumed that two teams have the same chances of winning a single game. For this exercise, assume that Team \(\mathrm{A}\) has a 60\(\%\) chance of defeating Team \(\mathrm{B}\) in every game they play against each other. a. Suppose there is a one-game tournament between the teams and the winner of the game wins the tournament. What is the probability that Team \(\mathrm{A}\) will win? That Team \(\mathrm{B}\) will win? b. Use a tree diagram to show all the possibilities for the tournament. For example, in the first game, there are two branches: A wins or \(\mathrm{B}\) wins. (Hint: If \(\mathrm{A}\) wins the first two games, is a third game played?) c. Suppose the teams played \(1,000\) tournaments. In how many tournaments would you expect Team \(\mathrm{A}\) to win the first game? In how many of those tournaments would you expect Team \(\mathrm{A}\) to also win the second game? d. For each combination in your tree diagram, use similar reasoning to find the number of tournaments out of \(1,000\) you would expect to go that way. For example, one combination should be ABB; in how many tournaments out of \(1,000\) would you expect the winner to be \(\mathrm{A},\) then \(\mathrm{B}\) , and then \(\mathrm{B}\) ? (Hint. Check your answers by adding them; they should total to \(1,000 . )\) e. Find the total number of tournaments out of \(1,000\) in which each team wins the tournament. What is the probability that Team \(\mathrm{A}\) wins a tournament? f. Which tournament, one-game or best-two-out-of-three, is better for Team B?

Suppose the \(3-o f-7\) lottery game was modified so that after each number was selected, that number was placed back into the group before the next number was selected. In this way, a number could be repeated, meaning triples such as \(1-2-2\) and \(3-3-3\) would be possible. a. How many possible pairs are there for this modified game, assuming that order does matter? Explain. b. Since order really doesn't matter in this game, \(1-1-2,1-2-1,\) and \(2-1-1\) are all the same triple. So there are only 84 possible different triples. Are all of these different triples equally likely? Explain. c. If you choose one number triple for this modified game, what is the probability you will win. (Hint: There are three cases to consider.)

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Consider the line \(y=-5 x-7\) a. A second line is parallel to this line. What do you know about the equation of the second line? b. Write an equation for the line parallel to \(y=-5 x-7\) that passes through the origin. 6\. Write an equation for the line parallel to \(y=-5 x-7\) that crosses the \(y\) -axis at the point \((0,-2) .\) d. Write an equation for the line parallel to \(y=-5 x-7\) that passes through the point \((3,0) .\)

To win the 3 -of- 10 lottery game, players must match three numbers from 1 to 10 with those selected in a random drawing. Remember that order doesn't matter. a. How many possible triples are there in the 3 of \(-10\) lottery game? b. What are your chances of winning the 3 of \(-10\) lottery game?

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