Calculate the probability for 2 different digits in 4 cells.

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Discussion Overview

The discussion revolves around calculating the probability of having exactly two different digits from the set S={1,2,3,4} appearing in four cells. The focus is on combinatorial arrangements and probability calculations related to this scenario.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the problem and provides examples of valid and invalid configurations for the digits in the cells.
  • Another participant suggests counting configurations where one digit appears once and another appears three times, as well as configurations where both digits appear twice.
  • A third participant proposes a formula involving permutations of the digits but questions its correctness.
  • Another participant challenges the proposed formula, indicating it may not be correct.
  • A later reply offers a broader hint, suggesting a combinatorial approach to select the two digits and calculate arrangements for different cases of their occurrences.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the proposed mathematical approach and calculations, indicating that there is no consensus on the solution or method to calculate the probability.

Contextual Notes

Some assumptions about the arrangements and the definitions of "good" configurations are not fully articulated, and the mathematical steps proposed by participants remain unresolved.

yevi
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The set S={1,2,3,4}, there are 4 cells.
What is the probability for 2 different digits (from S) to appear in 4 cells?
Example:

1,1,1,2 =>good
3,4,4,3 =>good
2,4,2,2 =>good

1,2,3,4 =>not good
1,2,3,3 =>not good
1,1,1,1=>not good
 
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Count those where you have 1 of one element 3 of another, and then 2 each of two elements.
 
Well, the 3 elements:

(4*3*2*3)/4^4

4 possibilities to choose first number, 3 possibilities to choose second number,
2 possibilities to choose third number and 3 possibilities to choose last number that has to be one of previous 3.

Is it right ?
 
Doesn't seem at all right to me.
 
This appears to be a home task. Since you have not informed that you got the solution, I am giving a broader hint.
1st select the two to appear in the cells--- in 4C2 ways.
For each of such choice find the no. of possible arrangements for the following cases.
1/ both of them is arising twice
2/ A specific no. of them arising thrice and the other once (and multiply this by 2,why?)
 

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