Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The distribution of scores on dice. Suppose that you have \(n\) dice, each a different color, all unbiased and sixsided. (a) If you roll them all at once, how many distinguishable outcomes are there? (b) Given two distinguishable dice, what is the most probable sum of their face values on a given throw of the pair? (That is, which sum between 2 and 12 has the greatest number of different ways of occurring?) (c) What is the probability of the most probable sum?

Short Answer

Expert verified
The number of distinguishable outcomes is \(6^n\). The most probable sum for two distinguishable dice is 7, with a probability of \( \frac{1}{6} \).

Step by step solution

01

Calculate the total number of distinguishable outcomes

Each die has 6 faces, and they are all different colors. Therefore, for each roll, there are 6 possible outcomes for each die. If there are n dice, multiply the number of possible outcomes for each die: Total outcomes = \(6^n\).
02

Determine the most probable sum for two dice

When calculating the possible sums of the face values between 2 and 12, note each individual sum's different combinations: Sum = 2: (1,1) Sum = 3: (1,2), (2,1) Sum = 4: (1,3), (2,2), (3,1) Continue this until all sums up to 12 have been considered. The sum with the most combinations is the most probable sum.
03

Find the most probable sum

Count the number of combinations for each sum: Sum = 2: 1 combination Sum = 3: 2 combinations Sum = 4: 3 combinations Sum = 5: 4 combinations Sum = 6: 5 combinations Sum = 7: 6 combinations Sum = 8: 5 combinations Sum = 9: 4 combinations Sum = 10: 3 combinations Sum = 11: 2 combinations Sum = 12: 1 combination The most probable sum is 7 because it has the most combinations – 6.
04

Calculate the probability of the most probable sum

Firstly, identify the total number of outcomes when rolling two dice: Total outcomes = \(6 \times 6 = 36\). Then use the number of combinations for the sum of 7: Probability = \( \frac{6}{36} = \frac{1}{6} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

probability distribution
Probability distribution refers to the way in which probability is spread over different possible outcomes of a random experiment. In the context of rolling dice, we are interested in how likely each sum of the dice values is. When rolling two six-sided dice, each face has an equal chance (1/6) of landing face-up. This creates a distribution of probabilities across these sums. For example, the sum of 2 has only one combination (1-1), resulting in a lower probability. On the other hand, the sum of 7 has six combinations (1-6, 2-5, 3-4, etc.), making it the most probable sum. Understanding this distribution helps predict outcomes and determine the most likely results in similar experiments.
dice outcomes
When we talk about dice outcomes, we refer to all the possible results of rolling a set of dice. For a single six-sided die, there are 6 possible outcomes. Rolling multiple dice increases the number of possible outcomes exponentially. For n dice, there are \(6^n\) possible outcomes. For example, with two dice, the number of outcomes is \(6 \times 6 = 36\). These outcomes can be represented as ordered pairs, where the first element is the outcome of the first die and the second element is the outcome of the second die. If the dice are distinguishable by color, each combination remains unique. This is important for calculating probabilities and understanding the overall behavior of the rolls.
combinatorial analysis
Combinatorial analysis involves counting the number of ways certain events can occur. In the case of two dice, we're interested in how many ways each possible sum can be made. This involves listing all pairs of outcomes (from 1-1 to 6-6) and noting which sums they produce. For example, the sum of 7 can be achieved in 6 different ways: (1-6), (2-5), (3-4), (4-3), (5-2), and (6-1). Combinatorial analysis helps us determine that certain sums (like 7) are more probable because there are more combinations that result in that sum. This technique simplifies understanding the probability distribution across all potential outcomes.
most probable sum
The most probable sum is the sum that occurs the most frequently when rolling a set of dice. For two six-sided dice, the most probable sum is 7. This is because there are more combinations of dice rolls that result in 7 than any other number. Specifically, there are six combinations: (1-6), (2-5), (3-4), (4-3), (5-2), and (6-1). To find the probability of this sum, divide the number of favorable combinations by the total number of outcomes. In this case, \(\frac{6}{36} = \frac{1}{6}\). Understanding the most probable sum helps predict the outcomes of rolling dice, which is useful in games and probabilistic modelling.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Probabilities of sequences. Assume that the four bases A, C, T, and G occur with equal likelihood in a DNA sequence of nine monomers. (a) What is the probability of finding the sequence AAATCGAGT through random chance? (b) What is the probability of finding the sequence AAAAAAAAA through random chance? (c) What is the probability of finding any sequence that has four A's, two T's, two G's, and one C, such as that in (a)?

Ion-channel events. A biological membrane contains \(N\) ion-channel proteins. The fraction of time that any one protein is open to allow ions to flow through is q. Express the probability \(P(m, N)\) that \(m\) of the channels will be open at any given time.

Presidential election. Two candidates are running for president. Candidate \(A\) has already received \(80 \mathrm{elec}-\) toral votes and only needs 35 more to win. Candidate \(B\) already has 50 votes, and needs 65 more to win. Five states remain to be counted. Winning a state gives a candidate 20 votes; losing gives the candidate zero votes. Assume both candidates otherwise have equal chances to win in those five states. (a) Write an expression for \(W_{A}\), total, the number of ways A can succeed at winning 40 more electoral votes. (b) Write the corresponding expression for \(W_{B, \text { total. }}\). (c) What is the probability candidate \(A\) beats candidate \(B\) ?

DNA synthesis. Suppose that upon synthesizing a molecule of DNA, you introduce a wrong base pair, on average, every 1000 base pairs. Suppose you synthesize a DNA molecule that is 1000 bases long. (a) Calculate and draw a bar graph indicating the yield (probability) of each product DNA, containing 0,1 , 2 , and 3 mutations (wrong base pairs). (b) Calculate how many combinations of DNA sequences of 1000 base pairs contain exactly 2 mutant base pairs. (c) What is the probability of having specifically the 500 th base pair and the 888 th base pair mutated in the pool of DNA that has only two mutations? (d) What is the probability of having two mutations side-by-side in the pool of DNA that has only two mutations?

A pair of aces. What is the probability of drawing two aces in two random draws without replacement from a full deck of cards?

See all solutions

Recommended explanations on Combined Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free