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Which is more likely: rolling a total of 9 when two dice are rolled or rolling a total of 9 when three dice are rolled?

Short Answer

Expert verified
Rolling a total of 9 with three dice is more likely.

Step by step solution

01

Determine possible outcomes for two dice

The total number of outcomes when rolling two dice is calculated by multiplying the number of faces on each die: \( 6 \times 6 = 36 \).
02

Calculate favorable outcomes for rolling a 9 with two dice

List all pairs of dice rolls that sum to 9: \((3,6), (4,5), (5,4), (6,3)\). There are 4 favorable outcomes.
03

Determine probability for two dice

Probability is the number of favorable outcomes divided by the total number of possible outcomes: \( P(9 \text{ with 2 dice}) = \frac{4}{36} = \frac{1}{9} \).
04

Determine possible outcomes for three dice

The total number of outcomes when rolling three dice is calculated by multiplying the number of faces on each die: \( 6 \times 6 \times 6 = 216 \).
05

Calculate favorable outcomes for rolling a 9 with three dice

List all combinations of three dice rolls that sum to 9: \((1,2,6), (1,3,5), (1,4,4), (2,2,5), (2,3,4), (3,3,3), etc.\). There are 25 favorable outcomes.
06

Determine probability for three dice

Probability is the number of favorable outcomes divided by the total number of possible outcomes: \( P(9 \text{ with 3 dice}) = \frac{25}{216} \).
07

Compare probabilities

Compare the probabilities calculated in earlier steps: \(P(9 \text{ with 2 dice}) = \frac{1}{9} \) which is approximately 0.111 and \(P(9 \text{ with 3 dice}) = \frac{25}{216} \) which is approximately 0.116. Since 0.116 > 0.111, rolling a 9 with three dice is more likely.

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Key Concepts

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

Dice Probability
When talking about dice probability, we are dealing with the likelihood of a particular outcome from rolling dice, which are fair six-sided objects. Each side is numbered from 1 to 6 and has an equal chance of landing face up. This is what we call a uniform distribution.
When calculating probability with dice, an important part is the total number of possible outcomes. For example, the probability of rolling a specific number (like a 3) on a single die is \ \( P = \frac{1}{6} \ \) because there are six faces and only one face with the number 3.
When multiple dice are involved, we need to consider all the possible combinations of numbers that can appear. This is because each die operates independently of the others.
For example:
  • Rolling two dice gives us 36 different combinations (6 sides per die, so \ \( 6 \times 6 = 36 \ \)).
  • Rolling three dice gives us 216 different combinations (6 sides per die, so \ \( 6 \times 6 \times 6 = 216 \ \)).
Favorable Outcomes
Favorable outcomes refer to the specific outcomes we are interested in. In this problem, we are looking at the outcomes that sum up to 9. To find these,
we list all possible ways the dice can add up to this number. For two dice:
  • (3,6)
  • (4,5)
  • (5,4)
  • (6,3)

These four pairs are the favorable outcomes when rolling two dice.
For three dice:
  • (1,2,6)
  • (1,3,5)
  • (1,4,4)
  • (2,2,5)
  • (2,3,4)
  • (3,3,3)

These are part of the 25 favorable outcomes when rolling three dice.
Finding favorable outcomes is crucial as it directly affects the probability calculation.
Combinatorial Analysis
Combinatorial analysis involves counting the number of ways events can occur. This helps us determine the total number of possible outcomes and the favorable ones. The product rule is often used here, which states that if one event can occur in m ways and a second event can occur independently of the first in n ways, then the two events can occur in \ \( m \times n \ \) ways.
In the context of dice rolling:
  • The total outcomes for two dice are \ \( 6 \times 6 = 36 \ \).
  • The total outcomes for three dice are \ \( 6 \times 6 \times 6 = 216 \ \).

For favorable outcomes, combinatorial analysis helps us list and count the specific combinations that lead to our target sum (like 9). This detailed examination ensures we don't miss any possibilities.
Total Outcomes
Total outcomes refer to the complete set of possible results when an event occurs. For dice rolling, each die has six faces, and so the outcomes multiply based on the number of dice rolled.
Here are some key points:
  • For one die, the total number of outcomes is 6.
  • For two dice, the total number of outcomes is \ \( 6 \times 6 = 36 \ \).
  • For three dice, the total number of outcomes is \ \( 6 \times 6 \times 6 = 216 \ \).

Knowing the total outcomes is essential for calculating probabilities. It forms the denominator in the probability fraction: \ \( P(x) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} \ \). For example, determining the likelihood of rolling a total of 9 with two dice involves dividing the number of favorable pairs (4) by the total possible outcomes (36). Thus, \ \( P(9 \text{ with 2 dice}) = \frac{4}{36} = \frac{1}{9} \ \). Similarly, for three dice, \ \( P(9 \text{ with 3 dice}) = \frac{25}{216} \ \).

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Most popular questions from this chapter

What is the probability of these events when we randomly select a permutation of the 26 lowercase letters of the English alphabet? a) The first 13 letters of the permutation are in alphabetical order. b) \(a\) is the first letter of the permutation and \(z\) is the last letter. c) \(a\) and \(z\) are next to each other in the permutation. d) \(a\) and \(b\) are not next to each other in the permutation. e) \(a\) and \(z\) are separated by at least 23 letters in the permutation. f) \(z\) precedes both \(a\) and \(b\) in the permutation.

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Use Chebyshev's inequality to find an upper bound on the probability that the number of tails that come up when a biased coin with probability of heads equal to 0.6 is tossed \(n\) times deviates from the mean by more than \(\sqrt{n}\) .

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In Exercises 18, 20, and 21 assume that the year has 366 days and all birthdays are equally likely. In Exercise 19 assume it is equally likely that a person is born in any given month of the year. a) What is the probability that two people chosen at random were born during the same month of the year? b) What is the probability that in a group of n people chosen at random, there are at least two born in the same month of the year? c) How many people chosen at random are needed to make the probability greater than 1?2 that there are at least two people born in the same month of the year?

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