Probability of Choosing E in ENERGISE: Combinatorics Approach

In summary, the conversation is discussing the probability of an E being one of the 4 randomly chosen letters from the word ENERGISE. The question is whether the probability is for EXACTLY 1 E or AT LEAST 1 E among the 4 letters. The answer provided by the book is incorrect, as confirmed by the teacher and the correct answer is 3/7.
  • #1
catalyst55
24
0
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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  • #2
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
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
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
No, you're not. I'm just losing it slowly... :biggrin:

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

1. What is combinatorics?

Combinatorics is a branch of mathematics that deals with counting and organizing objects or arrangements in a systematic way.

2. What are the basic principles of combinatorics?

The basic principles of combinatorics include the fundamental counting principle, permutations, combinations, and the inclusion-exclusion principle.

3. How is combinatorics used in real life?

Combinatorics has many real-life applications, such as in computer science, genetics, economics, and cryptography. It is used to solve problems related to arranging objects or making choices.

4. What is the difference between permutations and combinations?

Permutations are ordered arrangements of objects, while combinations are unordered selections of objects. In other words, the order of objects matters in permutations, but not in combinations.

5. How do I approach a simple combinatorics problem?

To solve a simple combinatorics problem, start by identifying the type of problem (permutations or combinations). Then, use the appropriate formula or principle to calculate the number of possible outcomes. It is also helpful to draw diagrams or make lists to visualize the problem.

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