What Are the Calculations Behind Probabilities in Combinatorial Problems?

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Discussion Overview

The discussion revolves around the calculations of probabilities in combinatorial problems, specifically focusing on two distinct problems involving digit combinations and symbol distributions. Participants explore the reasoning behind their approaches and seek clarification on the underlying principles of probability in these contexts.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a problem involving the probability of obtaining 1, 2, 3, or 4 different digits from a set of 4 digits, expressing confusion about the calculation for two different digits.
  • The same participant calculates the probability for one digit as 10/1000 and attempts to reason through the combinations for two different digits, arriving at a total of 45 possibilities but questions the correct answer of 630.
  • In the second problem, the participant seeks to find the probability of obtaining five symbols of each kind from a sample of size 25, initially proposing a calculation that is later identified as incorrect.
  • Another participant questions the interpretation of "two different digits," asking for clarification on whether a repeated digit counts as one or two different digits.
  • A subsequent reply clarifies that the interpretation is that repeated digits count as two different digits, suggesting that without this interpretation, achieving four different digits would be impossible.
  • Regarding the second question, a participant notes an unstated assumption of equal probability among the symbols and explains the probability of one specific sequence versus the total number of sequences with five of each letter.

Areas of Agreement / Disagreement

Participants express differing interpretations of the definitions of "different digits" and the assumptions underlying the probability calculations. There is no consensus on the correct interpretation or the calculations presented, indicating ongoing uncertainty and debate.

Contextual Notes

The discussion highlights potential limitations in the assumptions made about the distribution of digits and symbols, as well as the interpretations of the problems posed. The calculations presented depend on these interpretations, which remain unresolved.

Who May Find This Useful

This discussion may be useful for individuals interested in combinatorial probability, particularly those looking to understand the nuances of defining categories in probability problems and the implications of different interpretations on calculations.

lesdavies123
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Hi, there are two problems I can't really solve yet they don't seem that difficult. The two of them seem pretty related to me, I think there's something I'm not getting I'll detail my attempts at solving but any help especially with the steps to the solution would be really appreciated as I have the answers, these are not homework or anything like that just me practicing.

The first one is: Consider 4 digits, what are the probabilities of getting 1 different digit, 2 of them, 3 of them or 4.
I have no problem with 1 digit, it is 10/1000. The third I can do also and the 4th, it is especially with two different digits that I struggle.

So I thought the two digits which can repeat twice can be 0,1;0,2;0,3;0,4;0,5,etc...1,2;1,3;1,4;1,5..etc,2,3;2,4;2,5...etc,2,9; until 8,9 so a total of 45 possibilities.. but out of 10000? because 10^4? Anyways the answer is supposed to be 630... I don't know how that's possible.. Thanks

2nd problem

From the population of five symbols a, b, c, d, e, a sample of size 25 is taken. Find the probability that the sample will contain five symbols of each kind.

The only thing I could think of was (4/5)^5 x (3/5)^5 x (2/5)^5 x (1/5)^5 which is obviously wrong. Apparently the answer is 25!/(5!^5 X 5^25 ) so if anyone could explain the logic behind this answer would be great! Thank you guys! :)
 
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What, exactly, do you mean by "two different digits"? Different from what? If you number were "1101" would that count as "one different digit" (since the single digit "0" is different from the other digits) or would it count as "two different digits" (since it has "1" and "0" as digits)?
 
I had interpreted it as counting as 2 different digits! Or else 4 different would be impossible.
 
For your second question, there is an unstated assumption that all the letters are equally probable.
1/5^25 is the probability for one particular sequence of 25 letters.
25!/ 5!^5 is the number of sequences with 5 of each letter.
 

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