Draw a probability table for throwing two dice

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Homework Help Overview

The discussion revolves around calculating probabilities related to the sums obtained from throwing 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, Problem interpretation

Approaches and Questions Raised

  • Participants explore various interpretations of the problems, questioning the calculations and assumptions behind the probabilities provided. Some participants seek clarification on how the original poster derived their solutions, while others suggest alternative interpretations of the second problem.

Discussion Status

The discussion is active, with participants questioning the accuracy of the original poster's solutions and exploring different interpretations of the problems. Some guidance has been offered regarding the need for clarity in the assumptions made, particularly in relation to the second problem.

Contextual Notes

There is mention of specific sums and their properties, as well as the interpretation of factors and their sums. The original poster's solutions are being scrutinized, and there is a recognition of the complexity involved in determining the probabilities accurately.

npellegrino
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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. 7/36
3. 31/36

Any advice would be great. Thanks in advanced.
 
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hi npellogrino - do you have a specific question? if you want the thinking checked - how did you get your solutions?
 


I am questioning my solution for the 2nd problem, wouldn't it be 7/11 since there are only 11 possibile sums (2,3,4,5,6,7,8,9,10,11.12) and the factor's sums are only even for (3,5,6,7,10,11,12)?

I attached my work below.
 

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Also i believe the 3rd problem is 21/36
 


npellegrino said:
I am questioning my solution for the 2nd problem, wouldn't it be 7/11 since there are only 11 possibile sums (2,3,4,5,6,7,8,9,10,11.12) and the factor's sums are only even for (3,5,6,7,10,11,12)?

I attached my work below.

i find the question that a little difficult to follow,but i would interpret it as base prime factors and not include 1...
sum facs fac sum
2 2 2 (even)
3 3 3 (odd)
4 2.2 4 (even)
5 5 5 (odd)
6 3.2 5 (odd)... and so on though it is open to interpretation...

in any case for the total probability you must use you matrix to determine occurrence, for above
2 has a 1/36 probability
4 has a 3/36 probability...

so the total probability will be
P(tot) = P(2) + P(4) +...
 
Last edited:

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