Probability of Choosing E in ENERGISE: Combinatorics Approach

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Discussion Overview

The discussion revolves around calculating the probability of selecting the letter 'E' from the word ENERGISE when choosing 4 letters at random. Participants explore different interpretations of the question, specifically whether it asks for the probability of having exactly one 'E' or at least one 'E' among the chosen letters.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant calculates the probability of selecting exactly one 'E' using combinatorial methods, arriving at a different answer than the book.
  • Another participant questions the book's answer of 55/56, suggesting that the calculation may be based on a different interpretation of the problem.
  • A participant identifies two possible interpretations of the question: (i) the probability of exactly one 'E' and (ii) the probability of at least one 'E'.
  • There is a correction regarding the calculation of combinations, with one participant acknowledging a mistake in their arithmetic.
  • A later reply confirms the interpretation of the question as asking for exactly one 'E', asserting that the book's answer is incorrect.

Areas of Agreement / Disagreement

Participants express disagreement regarding the correct interpretation of the question and the validity of the book's answer. There is no consensus on the correct probability calculation, and multiple interpretations remain unresolved.

Contextual Notes

Participants note the importance of the exact wording of the question, which affects the probability calculations. There are unresolved mathematical steps and assumptions regarding the interpretations of the problem.

catalyst55
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The question goes something like this...

What is the probability of an E being one of the 4 randomly chosen letters from the word ENERGISE?

This is how i did it (the book says its wrong):

ENERGISE ==> 3 Es, 5 Non-Es (partitioning)

hence: (3c1*5c3)/(8c4)

the book has 55/56...

Cheers
 
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catalyst55 said:
the book has 55/56...

I don't see how they came up with that answer...I think your answer is correct.

(note 8C3 is 56...why would the book have that?)
 
Catalyst : I suggest you write down the question exactly as it appears. There are two similar looking possibilities : (i) the probability that there is EXACTLY 1 E among 4 randomly chosen letters , and (ii) the probability that there is AT LEAST 1 E among the 4 letters.

PS : You have answered assuming the first case. If that is the right case, then your answer is correct. Notice that the book's answer = 1 - your answer. If it is the second case, the book's answer is still wrong.
 
Gokul43201 said:
Notice that the book's answer = 1 - your answer.


Why do you say that? catalyst55's answer is 3/7...1-3/7=4/7. :confused: Am I missing something here? :redface:
 
No, you're not. I'm just losing it slowly... :biggrin:

My bad there. I must have forgotten how to multiply ! :redface:
 
Gokul43201 said:
Catalyst : I suggest you write down the question exactly as it appears. There are two similar looking possibilities : (i) the probability that there is EXACTLY 1 E among 4 randomly chosen letters , and (ii) the probability that there is AT LEAST 1 E among the 4 letters.

Hey Gokul43201,

I've thought about that and I've concluded that its definitely the former -- either way the answer on the back is wrong.

I've also asked my teacher and he's confirmed this.

Thanks a lot for your help guys.
 

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