simple combinatorics q

[catalyst55]

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

[Townsend]

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?)

[Gokul43201]

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.

[Townsend]

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:

[Gokul43201]

No, you're not. I'm just losing it slowly... :biggrin:

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

[catalyst55]

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 ive 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.