Altered die problem (Probability)

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Discussion Overview

The discussion revolves around a probability problem involving a bag containing two fair dice and one altered die. The altered die has all odd numbers replaced with "1". Participants are exploring the probability that the selected die is the altered die given that two rolls resulted in "1". The scope includes mathematical reasoning and probability calculations.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant presents the problem and seeks assistance in calculating the probability.
  • Another participant asks for clarification on what calculations have been performed so far.
  • A participant inquires about the probabilities of rolling "1" twice with both the regular and altered dice.
  • A later reply provides a calculation using Bayes' theorem, proposing that P(A|B) = (1/12)/(5/9) = 3/20, and asks if this is correct or if there are alternative approaches.

Areas of Agreement / Disagreement

The discussion does not appear to reach a consensus on the correctness of the calculations or the approach to the problem, as participants are still exploring different aspects and methods.

Contextual Notes

There are unresolved assumptions regarding the calculations and the definitions of probabilities used in the discussion. The participants have not fully clarified the steps leading to their conclusions.

bobsz
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Hello everyone,

I need some help with the following prob:

A bag contains 3 dice, 2 fair and 1 altered with all odd numbers replace with "1". One die is randomly selected and rolled independently twice. If the outcomes of both rolls were "1" and "1", what is the prob that the selected die is the altered die?

Thanks
 
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What have you done?
 
What is the probability of rolling "1" two times in a row with a regular die? What is the probability of rolling "1" two times in a row with an altered die?
 
Altere die problem

Here's what I've done:

Let A= Altered die and B=outcome is 1

then P(A|B) = P(A∩B)/ P(B)
P(A∩B) = 1/12 ( outcome of 3 11's)
P(B)= 5/9 ( total outcome of 1's)

therefore P(A|B) = (1/12)/(5/9) = 3/20

Is this correct? Is there another way to approach this problem?

Thanks!
 

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