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# What is the probability of randomly selecting an arrangement

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Senior Manager
Joined: 21 Oct 2013
Posts: 442
What is the probability of randomly selecting an arrangement [#permalink]

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20 Jul 2014, 10:50
2
KUDOS
14
This post was
BOOKMARKED
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Difficulty:

45% (medium)

Question Stats:

72% (02:01) correct 28% (02:34) wrong based on 162 sessions

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What is the probability of randomly selecting an arrangement of the letters of the word "MEDITERRANEAN" in which the first letter is E and the last letter is R?

A. 1/13
B. 1/20
C. 1/26
D. 1/50
E. 1/100

[Reveal] Spoiler:
13 letters, of which;
E = 3
N = 2
A = 2
R = 2

Hi, can we say Probability for E * Probability for R = 3/13 * 2/12 ?
It still end up same answer. Could be coincidence.
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 44288
Re: What is the probability of randomly selecting an arrangement [#permalink]

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20 Jul 2014, 11:01
3
KUDOS
Expert's post
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This post was
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goodyear2013 wrote:
What is the probability of randomly selecting an arrangement of the letters of the word "MEDITERRANEAN" in which the first letter is E and the last letter is R?

A. 1/13
B. 1/20
C. 1/26
D. 1/50
E. 1/100

[Reveal] Spoiler:
13 letters, of which;
E = 3
N = 2
A = 2
R = 2

Hi, can we say Probability for E * Probability for R = 3/13 * 2/12 ?
It still end up same answer. Could be coincidence.

There are 13 letters in MEDITERRANEAN, out of which E appears thrice, N appears twice, A appears twice, and R appears twice.

The number of arrangements of MEDITERRANEAN is therefore $$\frac{13!}{3!2!2!2!}$$.

The number of arrangements in which the first letter is E and the last letter is R is $$\frac{11!}{2!2!2!}$$ (we are left with 11 letters out of which E appears twice, N appears twice and A appears twice).

$$P = \frac{(\frac{11!}{2!2!2!})}{(\frac{13!}{3!2!2!2!})}=\frac{11!}{2!2!2!}*\frac{3!2!2!2!}{13!}=\frac{3!}{12*13}=\frac{1}{26}$$.

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What is the probability of randomly selecting an arrangement [#permalink]

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13 Jul 2015, 13:12
reto wrote:
What is the probability of randomly selecting an arrangement of the letters of the word "MEDITERRANEAN" in which the first letter is E and the last letter is R?

A. 1/13
B. 1/20
C. 1/26
D. 1/50
E. 1/100

Is this a tough combinations/probability question? I have no difficulty level. I find it hard.

Dont know about the difficulty but if you adopt 1 method to deal with arrangements, you can solve all the questions once you realise what is getting asked.

We have the word mediterranean which has 13 letters out of which E is 3, R/A/N are 2 each.

Again, probability = favorable outcomes / total outcomes

Total outcomes = arrangement of 13 letters with 3 Es and 2 each of R/A/N = 13! / (3!*2!*2!*2!)

Favorable outcomes = E <<11 letters>> R. Now E can be selected in 1 way out of 3 Es as all are same . Similarly R can be selected in 1 way out of 3 Rs as all Rs are same.

Thus we have : 1 (for E) <<11 letters with 2 Es , 2 A/N remaining while rest of them are 1 each>> 1 = 1* 11!/(2!*2!*2!) * 1

Dividing Favorable outcomes by Total outcomes , we get ----> $$\frac{1* 11!/(2!*2!*2!) * 1}{13! / (3!*2!*2!*2!)}$$ = 1/26. Thus B is the correct answer.
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Location: Switzerland
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Re: What is the probability of randomly selecting an arrangement [#permalink]

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25 Jul 2015, 05:56
4
KUDOS
goodyear2013 wrote:
What is the probability of randomly selecting an arrangement of the letters of the word "MEDITERRANEAN" in which the first letter is E and the last letter is R?

A. 1/13
B. 1/20
C. 1/26
D. 1/50
E. 1/100

[Reveal] Spoiler:
13 letters, of which;
E = 3
N = 2
A = 2
R = 2

Hi, can we say Probability for E * Probability for R = 3/13 * 2/12 ?
It still end up same answer. Could be coincidence.

Why do you calculate the total combinations and everything, it is much easier to calculate as follows:

3/13 * 1 * 1 * 1 * 1 ..... * 2/12 = 6/156 = 1/26

This is because 3/13 = Probability that first letter will be E and 2/12 = probability that the first letter will be R. Between everything else cuts down to 1*1*1... e.g. probability of the second letter is 11/11 = 1, > we do not care what's in between.

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Math Expert
Joined: 02 Aug 2009
Posts: 5713
Re: What is the probability of randomly selecting an arrangement [#permalink]

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25 Jul 2015, 06:07
Engr2012 wrote:
reto wrote:
What is the probability of randomly selecting an arrangement of the letters of the word "MEDITERRANEAN" in which the first letter is E and the last letter is R?

A. 1/13
B. 1/20
C. 1/26
D. 1/50
E. 1/100

Is this a tough combinations/probability question? I have no difficulty level. I find it hard.

Dont know about the difficulty but if you adopt 1 method to deal with arrangements, you can solve all the questions once you realise what is getting asked.

We have the word mediterranean which has 13 letters out of which E is 3, R/A/N are 2 each.

Again, probability = favorable outcomes / total outcomes

Total outcomes = arrangement of 13 letters with 3 Es and 2 each of R/A/N = 13! / (3!*2!*2!*2!)

Favorable outcomes = E <<11 letters>> R. Now E can be selected in 1 way out of 3 Es as all are same . Similarly R can be selected in 1 way out of 32 Rs as all Rs are same.

Thus we have : 1 (for E) <<11 letters with 2 Es , 2 A/N remaining while rest of them are 1 each>> 1 = 1* 11!/(2!*2!*2!) * 1

Dividing Favorable outcomes by Total outcomes , we get ----> $$\frac{1* 11!/(2!*2!*2!) * 1}{13! / (3!*2!*2!*2!)}$$ = 1/26. Thus B is the correct answer.

Hi,
i am sure you mean C.. 1/26
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Joined: 02 Aug 2009
Posts: 5713
Re: What is the probability of randomly selecting an arrangement [#permalink]

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25 Jul 2015, 06:12
reto wrote:
goodyear2013 wrote:
What is the probability of randomly selecting an arrangement of the letters of the word "MEDITERRANEAN" in which the first letter is E and the last letter is R?

A. 1/13
B. 1/20
C. 1/26
D. 1/50
E. 1/100

[Reveal] Spoiler:
13 letters, of which;
E = 3
N = 2
A = 2
R = 2

Hi, can we say Probability for E * Probability for R = 3/13 * 2/12 ?
It still end up same answer. Could be coincidence.

Why do you calculate the total combinations and everything, it is much easier to calculate as follows:

3/13 * 1 * 1 * 1 * 1 ..... * 2/12 = 6/156 = 1/26

This is because 3/13 = Probability that first letter will be E and 2/12 = probability that the first letter will be R. Between everything else cuts down to 1*1*1... e.g. probability of the second letter is 11/11 = 1, > we do not care what's in between.

Good one reto, i think you are bang on..
i dont find any flaw in the logic. ofcourse I too did it taking the favouable and total outcome..
kudos
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

GMAT online Tutor

Current Student
Joined: 20 Mar 2014
Posts: 2685
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: What is the probability of randomly selecting an arrangement [#permalink]

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25 Jul 2015, 06:24
reto wrote:
goodyear2013 wrote:
What is the probability of randomly selecting an arrangement of the letters of the word "MEDITERRANEAN" in which the first letter is E and the last letter is R?

A. 1/13
B. 1/20
C. 1/26
D. 1/50
E. 1/100

[Reveal] Spoiler:
13 letters, of which;
E = 3
N = 2
A = 2
R = 2

Hi, can we say Probability for E * Probability for R = 3/13 * 2/12 ?
It still end up same answer. Could be coincidence.

Why do you calculate the total combinations and everything, it is much easier to calculate as follows:

3/13 * 1 * 1 * 1 * 1 ..... * 2/12 = 6/156 = 1/26

This is because 3/13 = Probability that first letter will be E and 2/12 = probability that the first letter will be R. Between everything else cuts down to 1*1*1... e.g. probability of the second letter is 11/11 = 1, > we do not care what's in between.

reto, it is a matter of convenience. There is no 1 size fits all approach to quant (in general!). Your method is direct but it needs some time to understand and apply properly.

Additionally, the text in red above is not correct. I think you meant to write "last" letter will be R.

I also went by the favorable outcomes / total outcomes and was still able to solve it in 45 seconds. Ultimately, it all comes down to the easiest, fastest and correct (all these factors vary per individual) way that you are most comfortable way.
Intern
Joined: 07 Oct 2014
Posts: 2
Re: What is the probability of randomly selecting an arrangement [#permalink]

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28 Jul 2015, 12:18
Possibilities for 1st. letter: 3
Possibilities for 2nd. letter: 11
Possibilities for 3th. letter: 10
Possibilities for 4th. letter: 9
Possibilities for 5th. letter: 8
Possibilities for 6th. letter: 7
Possibilities for 7th. letter: 6
Possibilities for 8th. letter: 5
Possibilities for 9th. letter: 4
Possibilities for 10th. letter:3
Possibilities for 11th. letter: 2
Possibilities for 12th. letter: 1
Possibilities for 13th. letter: 2

So, total possibilities for words with 1st letter = E and the last letter = R: 11!*3*2

Total possibilities: 13!

So, the probability: (11!*3*2)/13! = (11!*3*2)/(13*12*11!) = (3*2)/(13*12) = 1/26
Intern
Joined: 13 Aug 2016
Posts: 2
Re: What is the probability of randomly selecting an arrangement [#permalink]

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11 Dec 2017, 00:33
Bunuel wrote:
goodyear2013 wrote:
What is the probability of randomly selecting an arrangement of the letters of the word "MEDITERRANEAN" in which the first letter is E and the last letter is R?

A. 1/13
B. 1/20
C. 1/26
D. 1/50
E. 1/100

[Reveal] Spoiler:
13 letters, of which;
E = 3
N = 2
A = 2
R = 2

Hi, can we say Probability for E * Probability for R = 3/13 * 2/12 ?
It still end up same answer. Could be coincidence.

There are 13 letters in MEDITERRANEAN, out of which E appears thrice, N appears twice, A appears twice, and R appears twice.

The number of arrangements of MEDITERRANEAN is therefore $$\frac{13!}{3!2!2!2!}$$.

The number of arrangements in which the first letter is E and the last letter is R is $$\frac{11!}{2!2!2!}$$ (we are left with 11 letters out of which E appears twice, N appears twice and A appears twice).

$$P = \frac{(\frac{11!}{2!2!2!})}{(\frac{13!}{3!2!2!2!})}=\frac{11!}{2!2!2!}*\frac{3!2!2!2!}{13!}=\frac{3!}{12*13}=\frac{1}{26}$$.

Hi Bunuel,

Since we have 3 E's and 2 R's isnt the number of ways to select E as first word - 3c1 and number of ways to select R - 2c1 ways?
Re: What is the probability of randomly selecting an arrangement   [#permalink] 11 Dec 2017, 00:33
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