How can I use probability to solve a dice problem with a two-dice table?

  • Thread starter Thread starter npellegrino
  • Start date Start date
  • Tags Tags
    Dice Probability
Click For Summary

Homework Help Overview

The discussion revolves around calculating probabilities related to the sums obtained from rolling two dice, specifically focusing on conditions such as divisibility by seven, the evenness of the sum of factors, and whether the sum is a composite number.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the construction of a probability table for two dice rolls and discuss the conditions for sums being divisible by seven, having even-summed factors, and being composite. There is uncertainty about whether to consider all possible sums or only specific ones.

Discussion Status

Some participants express confidence in certain solutions while others seek clarification on specific conditions, particularly regarding the interpretation of factors and the total number of outcomes considered. The conversation is ongoing with various interpretations being explored.

Contextual Notes

Participants are navigating the definitions of factors versus divisors and the implications of counting all possible outcomes from rolling two dice, which may affect their probability calculations.

npellegrino
Messages
17
Reaction score
0
Draw a table for throwing two dice (one red one blue). Find the probability.

1. That the sum is divisible by seven
2. The sum has factors whose sum is even
3. The sum is a composite number.

My solutions:

1. 1/6
2. 7/11
3. 7/12

Work is attached:
Any advice would be great. Thanks in advanced.
 

Attachments

  • work copy.jpg
    work copy.jpg
    29.4 KB · Views: 535
Physics news on Phys.org


npellegrino said:
Draw a table for throwing two dice (one red one blue). Find the probability.

1. That the sum is divisible by seven
2. The sum has factors whose sum is even
3. The sum is a composite number.

My solutions:

1. 1/6
2. 7/11
3. 7/12

Work is attached:
Any advice would be great. Thanks in advanced.

I believe #1 and #3 are correct. Still trying to get my head around #2, but you are probably correct.
 


Yes, 2 was tricky, it's the factors of the sum that are only even. But would I consider all 36 sums or only the 12 (2,3,4,5,6,7,8,9,10,11,12) hmmm
 


You need to consider all the 36 possible rolls, right? If I asked you, "what's the probability that a roll will come up 7?", you wouldn't answer 1/11, you'd answer 1/6 (I hope!). But I could phrase that question in many different ways: "What's the probability that the number on the bottom of one die is equal to the number on top of the other? What's the probability that you will roll the fourth prime number?", etc..

But no way of phrasing the question changes the fact that they want the *probability* that the condition will be fulfilled, which has to be weighted by how frequently the dice fall on the various numbers.
 


Take a look at the sum of 10. You have listed the factors 1,2,5,10. I would call those the divisors of 10. Had I been asked to factor 10 into prime factors I would have written 10 = 2*5. So the first thing to do is to be sure to know what interpretation of the question is intended.

Once you have determined which sums qualify for what you are looking for you can use the pdf of X = the sum of the dice to add P(X = k) for the qualifying values of k.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 53 ·
2
Replies
53
Views
10K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 11 ·
Replies
11
Views
4K
Replies
11
Views
2K
Replies
5
Views
4K
Replies
2
Views
3K
Replies
11
Views
3K