Is It Hypergeometric or Binomial: Calculating Guessing Probabilities on Exams?

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Discussion Overview

The discussion revolves around calculating the probability of correctly answering questions on a multiple-choice exam where each question has three possible answers. Participants explore whether the scenario should be modeled using hypergeometric or binomial probability distributions, specifically focusing on the probability of getting four or more answers correct by guessing.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant questions whether the situation is hypergeometric or binomial, given the structure of the exam.
  • Another participant argues that since each question allows for a correct or incorrect choice, it aligns with a binomial distribution.
  • A different participant presents two potential formulas for calculating the probability, suggesting one may be more appropriate than the other.
  • There is a clarification that the second formula is preferred, as it correctly reflects the probabilities without implying incorrect assumptions about the selection of answers.
  • One participant calculates a probability value based on the binomial approach, but expresses uncertainty about the accuracy of their calculation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the problem is best modeled by a hypergeometric or binomial distribution, and there is ongoing debate regarding the correct approach to calculating the probabilities.

Contextual Notes

Participants express uncertainty about the calculations and the assumptions underlying the choice of probability model. There are unresolved questions about the interpretation of the problem's parameters.

Somefantastik
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On a multiple guess exam, there are 3 possible answers for each of the 5 questions. What is the probability that the student will get four or more correct answers just by guessing?

Is this hypergeometric or binomial?
 
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Since on each question a student can either choose the correct answer or not, it is binomial. Do you see what is the probability a student will choose the correct answer on a specific question just by guessing?
 
I thought it was either this

[tex]\frac{(^{5}_{4}) (^{10}_{1})}{(^{15}_{5})}[/tex] choose 4 from 5 correct answers, choose 1 from 10 wrong answers divided by choosing 5 answers from 15

Or this (which I now know is probably the right answer)

[tex](^{5}_{4})p^{4}q + (^{5}_{5})p^{5}[/tex] which is the probability of 4/5 right answers + probability of 5/5 right answers, where prob correct answer = 1/3
 
Yes, the second is the one you want (you are NOT choosing 4 or 5 correct answers from all 15- that would imply that you could choose 2 correct answer from one problem and none from another!). Now what number is [tex](^{5}_{4})p^{4}q + (^{5}_{5})p^{5}[/tex]?
 
p = 1/3 and q = 2/3 => 0.14 unless I calculated wrong which is entirely possible.
 

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