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Chapter 2: Problem 42

You have three dice: one red (R), one green (G), and one blue (B). When all three dice are rolled at the same time, calculate the probability of the following out- comes: \(\mathbf{a} \cdot 6(\mathbf{R}), 6(\mathrm{G}), 6(\mathrm{B})\) b. \(6(\mathrm{R}), 5(\mathrm{G}), 6(\mathrm{B})\) c. \(6(\mathrm{R}), 5(\mathrm{G}), 4(\mathrm{B})\) d. No sixes at all e. A different number on all dice

Short Answer

Expert verified
a. \(\frac{1}{216}\), b. \(\frac{1}{216}\), c. \(\frac{1}{216}\), d. \(\frac{125}{216}\), e. \(\frac{5}{9}\).

Step by step solution

01

Calculate Probability for 6(R), 6(G), 6(B)

For each die, the probability of rolling a 6 is \( \frac{1}{6} \). Since the dice are independent, the probability of all three dice showing a 6 is the product of their individual probabilities. Therefore, the probability is \( \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216} \).
02

Calculate Probability for 6(R), 5(G), 6(B)

The probability of rolling a 6 on the red die is \( \frac{1}{6} \) and similarly \( \frac{1}{6} \) for the blue die. For the green die to roll a 5, the probability is also \( \frac{1}{6} \). Thus, the probability is \( \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216} \).
03

Calculate Probability for 6(R), 5(G), 4(B)

The probability for the red die to roll a 6 is \( \frac{1}{6} \), for the green die to roll a 5 is \( \frac{1}{6} \), and for the blue die to roll a 4 is \( \frac{1}{6} \). Multiply these probabilities to find the total probability: \( \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216} \).
04

Calculate Probability for No Sixes at All

The probability of not rolling a 6 on one die is \( \frac{5}{6} \). Since the dice are rolled independently, multiply the individual probabilities: \( \left( \frac{5}{6} \right)^3 = \frac{125}{216} \).
05

Calculate Probability for Different Numbers on All Dice

First, calculate the total possible outcomes, which is \( 6 \times 6 \times 6 = 216 \). For different numbers, choose a number for the first die (6 options), a different one for the second die (5 remaining options), and a different one for the third die (4 remaining options). The number of favorable outcomes is \( 6 \times 5 \times 4 = 120 \). Therefore, the probability is \( \frac{120}{216} = \frac{5}{9} \).

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Key Concepts

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

Dice Probability
Dice probability is a fundamental concept in probability theory, particularly when exploring outcomes from dice rolls. When you roll a fair die, each side has an equal chance of landing face up. For a standard six-sided die, each face has a probability of \(\frac{1}{6}\) since there are six outcomes.

In scenarios involving multiple dice, like our exercise, outcomes are analyzed by considering each die independently. This means you can calculate the probability of certain combinations, or outcomes, based on each die having six possible sides, each equally probable. Understanding dice probability helps lay the groundwork for more complex tasks involving probability calculations. The symmetry of a fair die ensures each roll is random and quite predictable in its structure.
Independent Events
Independent events is a key concept when dealing with probabilistic scenarios involving multiple actions, such as rolling multiple dice. Events are independent if the outcome of one does not affect the outcome of another. This is crucial when calculating the probability of combined events.

For example, in our exercise of rolling three dice, each die roll is independent. This is because the result of one die doesn鈥檛 influence the results of the others. Independence allows us to multiply the probabilities of individual events to find the probability of all the events occurring together.
  • If you roll one die and get a 6, it has no effect on rolling a 6 on another die.
  • You calculate the combined probability by multiplying the probabilities of each individual, independent event.
  • In probability theory, determining whether events are independent is crucial for accurate calculations.
Understanding the idea of independence is fundamental when you deal with multiple probabilistic events.
Combinatorics
Combinatorics relates to the counting of different arrangements and selections within a set. It plays a pivotal role in probability when determining how many ways an event can occur.

In our dice exercise, combinatorics helps in calculating how many outcomes possible when rolling three dice so that each die shows a different number. For example, with 6 possible numbers on each die, combinatorics helps in calculating that:
  • First die has 6 options.
  • Second die has 5 options (must be different from the first).
  • Third die has 4 options (different from the first two).
By multiplying these options, we find how many favorable outcomes exist. Knowing combinatorics is essential not only for probability but also in solving many problems where arrangement and selection are required.
Probability Calculations
Probability calculations are the methods used to determine the likelihood of different outcomes in an uncertain situation. Here, they involve several mathematical steps typically combining fundamental probability rules.

In our example, we perform probability calculations by considering both individual and joint probabilities. For each scenario, we identify the desired outcomes and use the rules of independent probability, such as:
  • Calculate using products for independent probabilities: the method for multiple events.
  • Apply combinatorial counts when permutations are important.
  • Consider complimentary probabilities (like the probability of 'not A') to simplify calculations.
By mastering probability calculations, you can solve real-world problems involving uncertainty, making this a vital mathematical skill.

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Most popular questions from this chapter

What is Mendel's first law?

A sex-linked recessive allele \(c\) produces a red-green color blindness in humans. A normal woman whose father was color blind marries a color-blind man. a. What genotypes are possible for the mother of the color-blind man? b. What are the chances that the first child from this marriage will be a color-blind boy? c. Of the girls produced by these parents, what proportion can be expected to be color blind? d. Of all the children (sex unspecified) of these parents, what proportion can be expected to have normal color vision?

Tay-Sachs disease (infantile amaurotic idiocy) is a rare human disease in which toxic substances accumulate in nerve cells. The recessive allele responsible for the disease is inherited in a simple Mendelian manner. For unknown reasons, the allele is more common in populations of Ashkenazi Jews of eastern Europe. A woman is planning to marry her first cousin, but the couple discovers that their shared grandfather's sister died in infancy of Tay-Sachs disease. a. Draw the relevant parts of the pedigree, and show all the genotypes as completely as possible. b. What is the probability that the cousins' first child will have Tay-Sachs disease, assuming that all people who marry into the family are homozygous normal?

Make up mnemonics for remembering the five stages of prophase I of meiosis and the four stages of mitosis.

Two black guinea pigs were mated and over several years produced 29 black and 9 white offspring. Ex plain these results, giving the genotypes of parents and progeny .
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