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Chapter 3: Problem 26

What is the probability of rolling two six-sided dice and obtaining the following numbers? a. 2 and 3 b. 6 and 6 c. At least one 6 d. Two of the same number (two \(1 \mathrm{s},\) or two \(2 \mathrm{s},\) or two \(3 \mathrm{s}\) etc.) e. An even number on both dice f. An even number on at least one die

Short Answer

Expert verified
a. \( \frac{1}{18} \), b. \( \frac{1}{36} \), c. \( \frac{11}{36} \), d. \( \frac{1}{6} \), e. \( \frac{1}{4} \), f. \( \frac{3}{4} \).

Step by step solution

01

Understand the Problem

We are tasked with finding the probability of certain outcomes with two six-sided dice. We need to calculate the probability for multiple scenarios, like getting specific numbers, having certain properties on one or both dice, and more.
02

Calculate Probability for Rolling a 2 and 3 (a)

With two dice, we can get a 2 and 3 in two different ways: (2,3) and (3,2). Since there are a total of 6 * 6 = 36 possible outcomes, the probability is \( \frac{2}{36} = \frac{1}{18} \).
03

Calculate Probability for Rolling a 6 and 6 (b)

Rolling two sixes corresponds to the single outcome (6,6). Thus, the probability is \( \frac{1}{36} \).
04

Calculate Probability for At Least One 6 (c)

First, calculate the probability of not rolling any 6. If neither die shows a 6, there are 5 options per die: \( 5 \times 5 = 25 \) combinations. Hence, the probability of at least one die showing a 6 is \( 1 - \frac{25}{36} = \frac{11}{36} \).
05

Calculate Probability for Two of the Same Number (d)

Each number from 1 to 6 can appear on both dice, leading to outcomes: (1,1), (2,2), ..., (6,6). Thus, we have 6 successful outcomes out of 36, so the probability is \( \frac{6}{36} = \frac{1}{6} \).
06

Calculate Probability for Even Numbers on Both Dice (e)

The even numbers on a die are 2, 4, and 6. Thus, there are 3 options per die leading to \( 3 \times 3 = 9 \) combinations, so the probability is \( \frac{9}{36} = \frac{1}{4} \).
07

Calculate Probability for an Even Number on At Least One Die (f)

First, find the probability of no even numbers, which means both dice showing odd values: 1, 3, or 5. That is \( 3 \times 3 = 9 \) outcomes, so \( \frac{9}{36} = \frac{1}{4} \). Thus, the probability for an even number on at least one die is \( 1 - \frac{1}{4} = \frac{3}{4} \).

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Key Concepts

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

Dice Probability
When we talk about dice probability, we're exploring how likely certain outcomes are when rolling dice. A standard die in probability theory is typically a six-sided cube numbered from 1 to 6. Understanding dice probability involves examining all possible outcomes and calculating the chance of landing on specific numbers or patterns. Dice probability is fundamental to grasping broader probability theory.
  • Each of the six sides on a six-sided die has an equal probability of being rolled, which is \( \frac{1}{6} \).
  • When rolling two dice, you multiply the outcomes, resulting in 36 possible combinations (since each die has 6 outcomes: 6 脳 6 = 36).
  • Common exercises involve calculating probabilities for outcomes like rolling doubles, getting a total sum, or rolling specific numbers.
These exercises are not only fun and engaging but also provide a basic understanding of how probability works.
Start with simple concepts before diving into the more complex scenarios like those involving conditional or independent probabilities.
Calculating Probabilities
Calculating probabilities requires a systematic approach to understanding all possible outcomes and what we consider a "successful" outcome.
Probability is defined as the number of successful outcomes divided by the total number of possible outcomes.
  • Let's consider rolling a 2 and 3 with two dice. There are two combinations: (2,3) and (3,2), so the probability is \( \frac{2}{36} = \frac{1}{18} \).
  • For scenarios where events can't happen simultaneously, like having at least one specific number appear, you use complement probabilities. In our exercise, to calculate the probability of rolling at least one 6, you first find out what the probability of not rolling a 6 is and subtract from 1.
These methods make probability calculations clearer and help in understanding how different scenarios affect the likelihood of events occurring.
With dice, always start by listing all possibilities to ensure complete analysis.
Outcomes of Dice Rolls
Understanding the outcomes of dice rolls involves knowing what results can be shown when dice are thrown.
Each time you roll a die, you have a series of possible results based on its number of sides.
  • For a single six-sided die, outcomes are 1, 2, 3, 4, 5, and 6.
  • When rolling two dice, the combined results are what's important. This could mean rolling identical numbers, such as doubles (1,1 up to 6,6), or specific combinations such as (2,3).
  • Different conditions like "rolling an even number" then rely on understanding which numbers fit the condition (such as 2, 4, 6) and counting the specific combinations that meet this requirement.
By accounting for all combinations, such as calculating the odds of rolling an even number on both dice versus at least one die, you can solve different probability questions accurately.
Outcomes form the backbone of probability calculations and are essential for predicting roll results.

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Most popular questions from this chapter

In sheep, lustrous fleece results from an allele \((L)\) that is dominant over an allele ( \(l\) ) for nomal fleece. A ewe (adult female) with lustrous fleece is mated with a ram (adult male) with normal fleece. The ewe then gives birth to a single lamb with normal fleece. From this single offspring, is it possible to detemine the genotypes of the two parents? If so, what are their genotypes? If not, why not? (IMAGE CANNOT COPY)

In mice, an allele for apricot eyes \((a)\) is recessive to an allele for brown eyes \(\left(a^{+}\right) .\) At an independently assorting locus, an allele for tan coat color \((t)\) is recessive to an allele for black coat color \(\left(t^{+}\right) .\) A mouse that is homozygous for brown eyes and black coat color is crossed with a mouse having apricot eyes and a tan coat. The resulting \(\mathrm{F}_{1}\) are intercrossed to produce the \(\mathrm{F}_{2}\). In a litter of eight \(\mathrm{F}_{2}\) mice, what is the probability that two will have apricot eyes and tan coats?

What is the principle of independent assortment? How is it related to the principle of segregation?

What characteristics of an organism would make it suitable for studies of the principles of inheritance? Can you name several organ isms that have these character istics?

In cats, curled ears result from an allele \((C u)\) that is dominant over an allele \((c u)\) for normal ears. Black color results from an independently assorting allele \((G)\) that is dominant over an allele for gray \((g) .\) A gray cat homozygous for curled ears is mated with a homozygous black cat with normal ears. All the \(F_{1}\) cats are black and have curled ears. a. If two of the \(F_{1}\) cats mate, what phenotypes and proportions are expected in the \(\mathrm{F}_{2} ?\) b. An \(F_{1}\) cat mates with a stray cat that is gray and possesses normal ears. What phenotypes and proportions of progeny are expected from this cross? (IMAGE CANNOT COPY)
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