curious perm/com probability question (1 Viewer)

[no_arg]

Here's a strange one!

a) A 4 letter word is to be constructed using the 8 characters EEEEEEGG.
In how many ways can this be done?

b) How many of the words in a) begin and end with an E?

c) What is the probability that a 4 letter word constructed from the 8 characters
EEEEEEGG begins and ends with an E?

 

[Riviet]

I'll get it started:

(a) 3 cases:

1) contains no G's = 1 way (EEEE)

2) contains 1 G = 4 ways (GEEE, EGEE, EEGE, EEEG)

3) contains 2 G's = 6 ways (GGEE, EGGE, EEGG, GEEG, GEGE, EGEG)

.'. total of 11 ways.

(b) Using (a), there are 4 of them.

Correct me if I'm wrong.

edit: correction from redruM (thanks)

 

[redruM]

I'll get it started:

(a) 3 cases:

1) contains no G's = 1 way (EEEE)

2) contains 1 G = 4 ways (GEEE, EGEE, EEGE, EEEG)

3) contains 2 G's = 4 ways (GGEE, EGGE, EEGG, GEEG)

.'. total of 9 ways.

(b) Using (a), there are 4 of them.

Correct me if I'm wrong.

What about GEGE or EGEG?

 

[pLuvia]

a)
Yeah I think 3) is:
If 2 G's and 2 E's
4!/(2!x2!)=6 ways
So there is a total of 11 ways

b)4 ways from a)

c)
P(constructing a word beginning and ending with E)=4/11

 

[no_arg]

!!!!!!!!!!!!!!!!!!!!!! But the answer is NOT 4/11 !!!!!!!!!!!!!!!!!!!!!!!

Imagine that I change the question so that the eight characters are replaced by 1000 E's and 2 G's.
You will still get 4/11 with the above attack but the correct answer is clearly very close to 1.

The problem is:

1) What is the correct probability for the original question
2) Why is the usual method failing!!

 

[no_arg]

But we almost always answer these questions via counting!!

An interesting point. If we ask the question

What is the probability that an 8 letter word constructed from the 8 characters
EEEEEEGG begins and ends with an E?

then both methods agree!

Why does the usual method of p(E)=n(E)/n(S) fail in the first instance??

 

[no_arg]

For the question

An eight letter word is constructed from EEEEEEGG. What is the prob it begins and ends with an E?

Method 1

Total number of eight letter words=8!/(2!6!)=28
Number that start and end with an E 6!/(2!4!)=15
Answer 15/28

Method 2 Prob(start E)xProb(end E)=6/8*5/7=30/56=15/28

The techniques are equivalent and both work just fine.

Why does method 1 fail for the original question??

 

[SeDaTeD]

The problem comes from what you are considering to be uniformly distributed.

You can consider each of the possible arrangements to have equal probability, ie. getting EE has the same probability as getting GG (from say EEEEEEGG).

The other method considers the letters being chosen from the list having equal probability, ie. getting EE has a larger probabiltiy than GG from the same list.

Constructing a word from a list would usually consider the latter case to be true.

p(E)=n(E)/n(S) does not work because the possible arrangements do not have equal probability if you consider the construction of the words to be done by taking letters from the list with equal probability.

The reason why constructing 8 letter words works with either method is that the list had 8 letters to start with. In this case the combinatorial and probabilistic problem are the same.

 

[no_arg]

But this is the question!
Why do they only coincide when n=8??

 

[SeDaTeD]

Because you chose the list to have 8 letters in the first place. Nothing special.

 

[AppleXY]

Combinatorics is a pain in the neck.... I had a similar question on my test wen I was doing the topic.. but now I can't remeber lol