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If you roll a pair of fair dice, what is the probability of (a) getting a sum of \(1 ?\) (b) getting a sum of \(5 ?\) (c) getting a sum of \(12 ?\)

Short Answer

Expert verified
(a) 0, (b) \(\frac{1}{9}\), (c) \(\frac{1}{36}\).

Step by step solution

01

Understanding the Dice Roll

When rolling two six-sided dice, each die has outcomes from 1 to 6. The total number of possible outcomes is calculated as the product of the outcomes of two dice, which is \(6 \times 6 = 36\).
02

Sum of 1

A sum of 1 is impossible because the smallest value each die can show is 1, making the smallest possible sum 2. Hence, there are 0 favorable outcomes for a sum of 1.
03

Probability of Sum of 1

The probability of an event is the number of favorable outcomes divided by the total number of possible outcomes. Here, \(P(\text{sum of 1}) = \frac{0}{36} = 0\).
04

Finding Favorable Outcomes for Sum of 5

To get a sum of 5, the dice can yield the following combinations: (1,4), (2,3), (3,2), and (4,1). This gives us 4 favorable outcomes.
05

Probability of Sum of 5

Using the formula, \(P(\text{sum of 5}) = \frac{4}{36} = \frac{1}{9}\).
06

Finding Favorable Outcomes for Sum of 12

To get a sum of 12, both dice must show a 6, which is the combination (6,6). There is only 1 favorable outcome.
07

Probability of Sum of 12

The probability is \(P(\text{sum of 12}) = \frac{1}{36}\).

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Key Concepts

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

Understanding the Dice Game
In the world of probability, the dice game is a classic example that brings together chance and mathematical calculations. When we talk about a dice game, we usually mean an activity involving one or more dice used to produce random outcomes. A common variant involves a pair of six-sided dice, each marked with numbers from 1 to 6. Each face of a die represents a different outcome, and the outcome of a roll is random.

When two dice are rolled together, numerous possibilities emerge. This randomness makes the dice game both exciting and intriguing, leading to various calculations and explorations in probability theory. Studying dice rolls helps us understand how chance works and is widely used for teaching basic principles of probability.

Whether in a game context or a mathematical exercise, analyzing the outcomes of dice rolls develops critical thinking skills and deepens comprehension of probability concepts.
Exploring Possible Outcomes
When considering possible outcomes in a dice game, it's essential to know the number of results you might get from a dice roll. In the case of rolling two dice, each die contributes 6 possible outcomes. This can be calculated by multiplying the outcomes of each die:
  • First Die: 6 outcomes (1, 2, 3, 4, 5, 6)
  • Second Die: 6 outcomes (1, 2, 3, 4, 5, 6)
  • Total Possible Outcomes: 6 \(\times\) 6 = 36


These 36 pairings represent every potential result when two dice are rolled simultaneously. Understanding this concept is crucial as it forms the basis for calculating the probability of any event involving the dice.

Each unique pairing of numbers (e.g., (1, 2) or (4, 5)) is considered a single possible outcome. Recognizing all possible outcomes allows you to explore and predict the probabilities of specific results, like getting a particular sum from the dice.
Identifying Favorable Outcomes
Favorable outcomes in a dice game are those results that satisfy the conditions of a specific event we are interested in. For example, if we're looking to achieve a sum of 5 when rolling two dice, we need to find which combinations of numbers meet that criterion.
  • Possible Rolls for Sum 5: (1, 4), (2, 3), (3, 2), (4, 1)
  • Number of Favorable Outcomes: 4


This concept is fundamental in probability because it helps us identify the exact instances where an event occurs out of all possible scenarios.

In a different case, if the event is to achieve a sum of 12, the only favorable outcome is (6, 6). Thus, there is only 1 favorable outcome.

Determining these favorable outcomes allows you to compute the probability of your event by comparing these to the total possible outcomes we discussed earlier.
Calculating Event Probability
Event probability refers to the likelihood of a specific outcome occurring in a given situation. It's calculated by dividing the number of favorable outcomes by the total number of possible outcomes. The formula is simple:

\[P( ext{event}) = \frac{ ext{Number of Favorable Outcomes}}{ ext{Total Number of Possible Outcomes}}\]

For example, to find the probability of rolling a sum of 5, we have 4 favorable outcomes (as identified earlier), and the total possible outcomes are 36.
  • Probability of Sum 5: \( P(\text{sum of 5}) = \frac{4}{36} = \frac{1}{9} \)


Similarly, for a sum of 12 where only one favorable outcome exists, the probability is:
  • Probability of Sum 12: \( P(\text{sum of 12}) = \frac{1}{36} \)


Understanding how to calculate event probability helps in analyzing various situations, assessing risks, and making informed decisions in both everyday life and complex mathematical situations.

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Most popular questions from this chapter

Imagine you have an urn containing 5 red, 3 blue, and 2 orange marbles in it. (a) What is the probability that the first marble you draw is blue? (b) Suppose you drew a blue marble in the first draw. If drawing with replacement, what is the probability of drawing a blue marble in the second draw? (c) Suppose you instead drew an orange marble in the first draw. If drawing with replacement, what is the probability of drawing a blue marble in the second draw? (d) If drawing with replacement, what is the probability of drawing two blue marbles in a row? (e) When drawing with replacement, are the draws independent? Explain.

Assortative mating is a nonrandom mating pattern where individuals with similar genotypes and/or phenotypes mate with one another more frequently than what would be expected under a random mating pattern. Researchers studying this topic collected data on eye colors of 204 Scandinavian men and their female partners. The table below summarizes the results. For simplicity, we only include heterosexual relationships in this exercise. \(^{42}\) $$\begin{array}{lcccc} & {\text { Partner (female) }} & \\ & \text { Blue } & \text { Brown } & \text { Green } & \text { Total } \\ \hline \text { Blue } & 78 & 23 & 13 & 114 \\ \text { Brown } & 19 & 23 & 12 & 54 \\ \text { Green } & 11 & 9 & 16 & 36 \\ \hline \text { Total } & 108 & 55 & 41 & 204\end{array}$$ (a) What is the probability that a randomly chosen male respondent or his partner has blue eyes? (b) What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes? (c) What is the probability that a randomly chosen male respondent with brown eyes has a partner with blue eyes? What about the probability of a randomly chosen male respondent with green eyes having a partner with blue eyes? (d) Does it appear that the eye colors of male respondents and their partners are independent? Explain your reasoning.

About \(30 \%\) of human twins are identical, and the rest are fraternal. Identical twins are necessarily the same sex - half are males and the other half are females. One-quarter of fraternal twins are both male, one-quarter both female, and one-half are mixes: one male, one female. You have just become a parent of twins and are told they are both girls. Given this information, what is the probability that they are identical?

The American Community Survey is an ongoing survey that provides data every year to give communities the current information they need to plan investments and services. The 2010 American Community Survey estimates that \(14.6 \%\) of Americans live below the poverty line, \(20.7 \%\) speak a language other than English (foreign language) at home, and \(4.2 \%\) fall into both categories. \(^{22}\) (a) Are living below the poverty line and speaking a foreign language at home disjoint? (b) Draw a Venn diagram summarizing the variables and their associated probabilities. (c) What percent of Americans live below the poverty line and only speak English at home? (d) What percent of Americans live below the poverty line or speak a foreign language at home? (e) What percent of Americans live above the poverty line and only speak English at home? (f) Is the event that someone lives below the poverty line independent of the event that the person speaks a foreign language at home?

Data collected at elementary schools in DeKalb County, GA suggest that each year roughly \(25 \%\) of students miss exactly one day of school, \(15 \%\) miss 2 days, and \(28 \%\) miss 3 or more days due to sickness. \(^{24}\) (a) What is the probability that a student chosen at random doesn't miss any days of school due to sickness this year? (b) What is the probability that a student chosen at random misses no more than one day? (c) What is the probability that a student chosen at random misses at least one day? (d) If a parent has two kids at a DeKalb County elementary school, what is the probability that neither kid will miss any school? Note any assumption you must make to answer this question. (e) If a parent has two kids at a DeKalb County elementary school, what is the probability that both kids will miss some school, i.e. at least one day? Note any assumption you make. (f) If you made an assumption in part (d) or (e), do you think it was reasonable? If you didn't make any assumptions, double check your earlier answers.

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