Question

Tough probability question: If you split up the word ELEMENT and put the letters back in a random order, what is the probability that the three Es are consecutive?

Answer 1

This was at MathCounts yesterday. I got an answer, but I'm not sure if it is right. I got 1/588

Answer 2

there are probably a couple different ways to do this, and probably some easier than this, but you can figure out the probability that they are picked first, that would be
\[\frac{3}{7}\times \frac{2}{6}\times \frac{1}{5}\]

Answer 3

That's not what I'm asking

Answer 4

\[\frac{1}{42}\]

Answer 5

yes, but then we can look at the possible placement of the three E
EEE X X X X
X EEE X X X
XX EEE X X
XXX EEE X
XXXX EEE

Answer 6

Suppose the three Es are first, i.e., we have
EEExxxx
How many such possibilities are there? 3! 4!

Now we can also have
xEEExxx, another 3! 4!
xxEEExx, 3!4!
xxxEEEx, 3!4!
xxxxEEE, 3!4!

for a total of 5 . 3!4!

Now divide that by the total number of possible arrangements.

Answer 7

since each of these has the same probability, and since each of these is different (disjoint events) then you can add them up

Answer 8

Wait a sec there, they are NOT the same
EEExxxx has a different probability than xxEEExx

Answer 9

But there are an equal number of possible arrangements
EEExxxx and xxEEExx

Answer 10

xxEEExx = x x 3/5 2/4 1/3 x x

Answer 11

as i said, probably several ways to do this. i get
\[5\times \frac{3}{7}\times \frac{2}{6}\times \frac{1}{5}\]
\[=\frac{1}{7}\] but i could certainly by wrong

Answer 12

Using my method I arrive at exactly the same answer:
\[ \frac{ 5 \cdot 3! 4!}{7!} = \frac{1}{7} \]

Answer 13

like is said, more ways that one to skin a cat

Answer 14

do you still consider the arrangements of the e's?

Answer 15

skin a cat? What I did was I knew that there was 5 possiblities of three consecutive e's

Answer 16

EEExxxx has a different probability than xxEEExx

\[\frac{3}{7}\times \frac{2}{6}\times \frac{1}{5}\times \frac{4}{4}\times \frac{3}{3}\times \frac{2}{2}\times \frac{1}{1}\]
same as
\[\frac{3}{7}\times \frac{2}{6}\times \frac{1}{5}\]

Answer 17

also, the total number of possibilities of letter combinations is not 7!, it's 7!/3! because there are 3 Es

Answer 18

In my calculation, I am considering the three Es to be distinguishable, in which case there are indeed 7! possibilities.

That is why is each of the scenarios
EEExxxx, etc.
I also count the 3! = 6 possible arrangements of the three Es.

Answer 19

but EEExxxx is not different from EEExxxx

Answer 20

Right, but I am counting correctly. That's all.

Answer 21

now we try
xxEEExx
\[\frac{4}{7}\times \frac{3}{6}\times \frac{3}{5}\times \frac{2}{4}\times \frac{1}{3}\times \frac{2}{2}\times \frac{1}{1}\]
think it is identical

Answer 22

I don't see where you're going james

Answer 23

In other words,
\[ E_1E_2E_3xxxx \] is not the same as
\[ E_3E_1E_2xxxx \]

But it is a possible arrangement. So I am counting up all of the possible arrangements of the seven letters which give three Es in a row. That is
\[ 5 \times 3!4! \]
then I divide that by the total number of possible arrangements of those seven letters, 7!

Answer 24

Oh, wow satellite, you are right!

Answer 25

when i did it i took the probability of xxEEExx as
1 1 3/5 2/4 1/3 1 1, but I didnt consider the chances the the first two letters WEREN'T Es

Answer 26

ah right, and to think even simpler, (bonehead that i am) once we compute the probability that we get three in a row first, clearly we can ignore what happens next, because once you get three in a row the probability everything else gives a 1, that is once you have 3 in a row on the first slot, you can ignore the other combinations

Answer 27

so with your thing satellite any 3-combo of the Es has a probability of
3/7 * 2/6 * 1/5 = 3/105

Answer 28

there are 5 different possible 3-combos of E, so multiply that by 5 to get 3/21 = 1/7, but then we have to divide it by 7!/3! because that is the total number of possible outcomes, correct?

Answer 29

It gives 1/5880

Answer 30

lol

Answer 31

i would say
\[\frac{3}{105}=\frac{1}{35}\] and that is the probability you get them first. then you have 5 possible placements, they are disjoint events, so multiply by 5 and get
\[\frac{1}{7}\]

Answer 32

Yes, the answer is 1/7

Answer 33

don't you still have to divide by the total number of outcomes?

Answer 34

the difference between our methods is i am not concerned with how many possible ways i can pick the seven letters. i am just computing one probability. no, i accounted for that.
if you want to divide by the number of possible outcomes, then use jamesj's method, which might make more sense

Answer 35

if you want a 7! in the denominator your numerator has to include all the possible ways to get 3 E in a row

Answer 36

Right. Sat has calculated the probability directly of each outcome
EEExxxx, etc.
and then added them.

I counted the number of such outcomes, EEExxxx etc., then divided it by the total number of possible outcomes.

Answer 37

I still don't understand.

Answer 38

Let me ask you: how many ways can the letters be chosen to be in order
EEELMNT

Answer 39

OH I see, James, you did 5 x 3! divided by (7!/3!) didnt you?

Answer 40

6 ways

Answer 41

No. I have
\[ \frac{5 3! 4!}{7!} \]

===

Yes 6 = 3!. Hence of all possible outcomes
EEExxxx there are

3! x 4! possbilities.

Answer 42

**correction: 5 . 3! 4! / 7!

Answer 43

what? where did you get 4!?

Answer 44

Because there are 4! ways of arraigning the L, M, N and T

Answer 45

*arranging

Answer 46

okay, so what is 5.

Answer 47

EEExxxx --> 3!4! possibilities
xEEExxx --> 3!4! possibilities
xxEEExx --> 3!4! possibilities
xxxEEEx --> 3!4! possibilities
xxxxEEE --> 3!4! possibilities

hence 5 x 3!4! possibilities in total

Answer 48

alright, then you divided by 7! why? I know there are seven letters, but if you want to find the total possible outcomes you have to divide by (number of repeated letters)! and so it would be 3!

Answer 49

No, there are 7! possible arrangements precisely because we counted the
EEELMNT as 6 possible arrangements, not 1.

Answer 50

If we counted EEELMNT as 1, then I would agree the total is 7!/3!

Answer 51

Oh, I see what you did, say we did count EEELMNT as 1, what would we do from there?

Answer 52

Then
EEExxxx
would have 4! possibilities
and in total
EEExxxx
...
xxxxEEE
would have
5 . 4! possibilities.

Answer 53

Then divide by...?

Answer 54

...and the probability would be
\[ \frac{ 5 \cdot 4!}{7!/3!} = \frac{ 5 \cdot 4! 3!}{7!} = \frac{1}{7} \]

Answer 55

Okay. Thanks. Now will you tell me what what is wrong with my way please? I have to go eat so take your time

Answer 56

I didn't see your method, I'm sorry. Post it again and when I'm back from answering other questions myself, I'll see if I can comment on it.

Answer 57

Alright. Well, I knew there was 5 different outcomes of 3 Es, so i found the probability of each like EEExxxx, xEEExxx, xxEEExx, xxxEEEx, xxxxEEE by doing 3/7 * 2/6 * 1/5 3/6 * 2/5 * 1/4 3/5 * 2/4 * 1/3 3/4 * 2/3 * 1/2 1 * 1 * 1, then I added the 5 probabilities together and divided by (7!/3!). I know I found each set of Es probability wrong. For example on xxEEExx i put 3/5 * 2/4 * 1/3 but it should've been 4/7 * 3/6 * 3/5 * 2/4 * 1/3. But that would've gotten me even further away from 1/7, my way I got 1/588, after fixing that mistake I got 1/5880 by doing (1/7)/(7!/3!). If the answer is 1/7, then how come I do not divide there?

Answer 58

I'm still confused what exactly is your method.

Answer 59

which part?

Answer 60

Ok, I think I see what's going on here now. Each of the probabilities you calculated is exactly that: a probability. Why are you then dividing them by some other number of outcomes. You have already normalized them; you have already put in a denominator in calculating probabilities. To do that again is unnecessary.

Answer 61

Oh, so the exact probability of xEEExxx is 4/7 * 3/6 * 2/5 * 1/4 * 1 * 1 * 1 not all of that divided by 7!/3!, that makes sense now. Thanks for all of your help James. That means I missed that question yesterday. Oh well ,we still got first place as a team.