Probability HELP (1 Viewer)

[StGeorgeDragons]

letters of word ENUMERATE are arranged in a line.


wats the prob that all the E's are seperated. can someone pls expalin this very simply and easily please. go through everystep, as if ur teaching someone who doesnt know wat probability means is.

 

[currysauce]

letters of word ENUMERATE are arranged in a line.



woooooooooooooo thanks then

 

[StGeorgeDragons]

the answer is, 6! x 7C3 x 3! / 9!


i think total is 9!, cause even when E's are swaped, the chances of those have to count.

 

[Dreamerish*~]

the answer is, 6! x 7C3 x 3! / 9!


i think total is 9!, cause even when E's are swaped, the chances of those have to count.

I guess you're asking the wrong two people.

Does the answer section explain how you get the answer?

Hmm, the 6! is the arrangement of the non-E letters, the 7C3 is the number of combinations of 3 letters from 7. Exactly which 7? And the 3! must be the number of arrangements for the E's.

 

[StGeorgeDragons]

it doesnt show how to get the answer

 

[serge]

the answer is

[All random combinations minus all combinations of 3Es together minus all combinations of 2Es together (as long as the 3rd E does not touch those 2Es)]
divided by all random combinations (since were looking for the probability)

I had trouble making sense of the 2Es not touching the 3rd E at first
but now i think that the expression should be equal to all possible 2E combinations minus all 3e combinations

 

[zozo]

if you are having trouble understanding the solution, here is an easier and more basic way to do it....

first, work out possible number of combinations of letters in ENUMERATE

9!
3! = 60480


then would have worked out the combination of all E's touching eg count EEE as 1
eg (EEE)NUMRAT

so the answer would be: 7! = 5040

then worked out the combination for 2 E's touching (count EE as 1)
eg (EE)NUMRATE

so the answer would be:
8!
2! =20160

so probability of the E's touching would be:

(5040+20160)/60480

THUS!

the probablity of NO E's touching would be:

P= 1 - (5040+20160)/60480
= 0.4166666666....

which you will find it the answer of that calculation

hope it helps... i find it easier when its broken up like that.

with their solution, you will find that they originally would have had

6! x 7C3
9!
3!

which, when you invert the bottom fraction and multply, becomes

6! x 7C3 x 3! / 9!

the bottom fraction of 9!/3! is jsut the total number of letter combinations (which i found in my first step)

hope that was simple enough for you

 

[StGeorgeDragons]

another part of the question is, wat is the prob that

exactly 2 of the E's are together?

 

[KFunk]

then worked out the combination for 2 E's touching (count EE as 1)
eg (EE)NUMRATE

so the answer would be:
8!
2! =20160

One thing I thought I'd mention is that the permutations with 2 E's touching also includes the cases where three E's are touching so if you take away the (EE) situation and the (EEE) then you're taking away the same thing twice.

 

[Dreamerish*~]

if you are having trouble understanding the solution, here is an easier and more basic way to do it....

first, work out possible number of combinations of letters in ENUMERATE

9!
3! = 60480


then would have worked out the combination of all E's touching eg count EEE as 1
eg (EEE)NUMRAT

so the answer would be: 7! = 5040

then worked out the combination for 2 E's touching (count EE as 1)
eg (EE)NUMRATE

so the answer would be:
8!
2! =20160

so probability of the E's touching would be:

(5040+20160)/60480

THUS!

the probablity of NO E's touching would be:

P= 1 - (5040+20160)/60480
= 0.4166666666....

which you will find it the answer of that calculation

hope it helps... i find it easier when its broken up like that.

with their solution, you will find that they originally would have had

6! x 7C3
9!
3!

which, when you invert the bottom fraction and multply, becomes

6! x 7C3 x 3! / 9!

the bottom fraction of 9!/3! is jsut the total number of letter combinations (which i found in my first step)

hope that was simple enough for you

OMG, dude, do you realise that's exactly what I did before, except for the bit in bold?

I feel like an ass.

 

[KFunk]

I was just thinking about it and I may have hit upon the way they got their answer.

Say you take the letters of NUMRAT. There are 6! ways you can arrange them in a line. If you arrange them in a line you get 7 spaces in between and on either side of the letters where you could place the E's ... like so:

₁N₂U₃M₄R₅A₆T₇

If you choose 3 of those 7 spaces (* there are ⁷C₃ ways to do this) and put the E's in them then they will be separated. This gives you 6!⁷C₃ ways arrangements where the E's are separated. Since there are 9!/3! total arrangements the the probability of the E's being separated is:

(6!⁷C₃)/(9!/3!) = (6!⁷C₃3!)/9!

 

[StGeorgeDragons]

oh thats cool, thanks

 

[KFunk]

another part of the question is, wat is the prob that

exactly 2 of the E's are together?

You can probably do this part in a similar fasion:

As before there are 6! ways you can arrange the letters of NUMRAT in a line. You then get 7 spaces...

₁N₂U₃M₄R₅A₆T₇

In this case you need to choose 2 of 7 spaces (⁷C₂ ways to do this) in which to put the EE and the E. Worth noting is that for each combination of 2 spaces there are 2 ways to arrange the EE and the E term between them. This gives us 2.6!⁷C₂ arrangements with exactly two EE's together. As before there are 9!/3! total arrangments making the total probability:

(2.3!6!⁷C₂)/9!


BTW: For this answer and the one above it would probably be more correct to say that there are 9! total arrangments and to multiply the number of arrangements with the E's seperated by 3! (likewise for the number of arrangements with the EE's together). It yields the same answer but as you said ealier: it's probability so you have to treat each E like a seperate entity (kinda). Also this means that the number of arrangments with EEE together is 7!3!. I just thought I'd mention this.