Simple combinatorics q

  • Thread starter catalyst55
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  • #1
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
 

Answers and Replies

  • #2
Townsend
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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?)
 
  • #3
Gokul43201
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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.
 
  • #4
Townsend
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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:
 
  • #5
Gokul43201
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No, you're not. I'm just losing it slowly... :biggrin:

My bad there. I must have forgotten how to multiply ! :redface:
 
  • #6
catalyst55
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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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