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by kanwar86 » Sat Sep 08, 2012 11:01 am
srivathsan wrote:If three letters are put in three envelope with three diff. addresses, what is the probability that no addressee receives the correct letter?

a.1/6
b.1/4
c.1/3
d.1/2
e.2/3

Kindly help me out
We have three different envelopes with addresses, and three letters.
Now, total number of ways in which letters can be put in envelopes is 3*2*1 = 6 ways
Let's see how to find the number of ways in which we can send at least one letter to correct address
If one is sent to correct address, the others must be sent to wrong addresses. This gives 3 ways.
If two are sent to correct addresses, this implies the third one will also be sent to correct address. This gives 1 such way.
Hence,there are 4 ways in which at least one letter will be sent to correct address.
So, number of ways in which no letter is put in the correct envelope = 6-4 = 2
So, the required probability = 2/6 = 1/3 (Option C)
Last edited by kanwar86 on Sat Sep 08, 2012 8:47 pm, edited 1 time in total.
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by GMATGuruNY » Sat Sep 08, 2012 7:39 pm
srivathsan wrote:If three letters are put in three envelope with three diff. addresses, what is the probability that no addressee receives the correct letter?

a.1/6
b.1/4
c.1/3
d.1/2
e.2/3

Kindly help me out
Let the correct ordering of the letters be ABC.

Arrangements in which no letter is in the correct position:
BCA
CAB
2 ways.

Total number of ways to arrange the letters = 3! = 6.

P(no letter is in the correct position) = 2/6 = 1/3.

For those who like formulas:

A DERANGEMENT is a permutation in which NO element is in the correct position.
Given n elements (where n>1):

Number of derangements = n! (1/2! - 1/3! + 1/4! + ... + ((-1)^n)/n!)

Given 3 letters A, B and C:
Total number of derangements = 3! (1/2! - 1/3!) = 3-1 = 2.
Total possible arrangements = 3! = 6.
P(no letter is in the correct position) = 2/6 = 1/3.

The correct answer is C.
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by Brent@GMATPrepNow » Sun Sep 09, 2012 6:58 am
srivathsan wrote:If three letters are put in three envelope with three diff. addresses, what is the probability that no addressee receives the correct letter?
a.1/6
b.1/4
c.1/3
d.1/2
e.2/3
Given the small number of possible outcomes here, we should also consider simply listing the possibilities.

Aside: Students often discount listing as a viable strategy because it doesn't seem very "mathematical," but we should always keep in mind that our goal here is not to impress our former math teachers; our goal is to maximize our GMAT score.

Okay, let's list the outcomes systematically. I'll list each outcome as a 3-element set consisting of a, b and c. The first element indicates the letter that Person A receives, the second element indicates the letter that Person B receives, and the third element indicates the letter that Person C receives.

So, for example, the outcome {b, a, c} indicates Person A getting letter b, Person B getting letter a, and Person C getting letter c. So, here one of the people (Person C) received the letter destined for him/her.

Okay, now for the outcomes

1. {a, b, c}
2. {a, c, b}
3. {b, a, c}
4. {b, c, a}
5. {c, a, b}
6. {c, b, a}

Of the 6 possible outcomes, only 2 of them are such that no person receives the correct letter.

So, P(no one gets the correct letter) = 2/6
= 1/3
= C

Cheers,
Brent
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by Manjareev » Sun Sep 16, 2012 5:19 am
i had never seen this formula. thanks.
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by das.ashmita » Mon Sep 17, 2012 3:08 am
My approach :

Probability (1st letter goes in wrong envelop) = 2/3
Probability (2nd letter goes in wrong envelop) = 1/2
Probability (3rd letter goes in wrong envelop) = 1

Therefore, P = 2/3 * 1/2 * 1 = 1/3
:)
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by Brent@GMATPrepNow » Mon Sep 17, 2012 7:38 am
das.ashmita wrote:My approach :

Probability (1st letter goes in wrong envelop) = 2/3
Probability (2nd letter goes in wrong envelop) = 1/2
Probability (3rd letter goes in wrong envelop) = 1

Therefore, P = 2/3 * 1/2 * 1 = 1/3
:)
This solution is somewhat problematic.

The first part, P(1st letter goes in wrong envelop) = 2/3, is correct.

However, how do we know that P(2nd letter goes in wrong envelop) = 1/2?
The assumption here is that the 1st letter did not go into the envelop intended for the 2nd letter. If the 1st letter did go into the envelop intended for the 2nd letter, then P(2nd letter goes in wrong envelop) = 1 (not 1/2).
This problematic issue also extends to the 3rd letter.
To deal with these issues, we need to consider a lot more cases than what your approach suggests.

Edit: I earlier suggested that this question may be out of scope for the GMAT. However, Mitch's slight adjustment to the solution (below)clears up the complexity issue.

Cheers,
Brent
Last edited by Brent@GMATPrepNow on Mon Sep 17, 2012 3:56 pm, edited 1 time in total.
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by GMATGuruNY » Mon Sep 17, 2012 2:54 pm
das.ashmita wrote:My approach :

Probability (1st letter goes in wrong envelop) = 2/3
Probability (2nd letter goes in wrong envelop) = 1/2
Probability (3rd letter goes in wrong envelop) = 1

Therefore, P = 2/3 * 1/2 * 1 = 1/3
:)
This approach is valid if we apply the following reasoning:

Let the correct ordering of the letters be A-B-C.
P(B or C is placed in the first envelope) = 2/3.

At this point, either B or C -- whichever of the two was NOT placed in the first envelope -- could still be placed in the correct corresponding envelope.
P(this letter is NOT placed in the correct envelope) = 1/2. (Of the 2 envelopes left, one is incorrect.)

Only one letter remains: A, which should have been placed in the first envelope.
P(A is not placed in the correct envelope) = 1.

To combine these probabilities, we multiply:
2/3 * 1/2 * 1 = 1/3.
Last edited by GMATGuruNY on Fri Sep 21, 2012 1:49 am, edited 1 time in total.
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by das.ashmita » Fri Sep 21, 2012 1:34 am
Brent@GMATPrepNow wrote:
das.ashmita wrote:My approach :

Probability (1st letter goes in wrong envelop) = 2/3
Probability (2nd letter goes in wrong envelop) = 1/2
Probability (3rd letter goes in wrong envelop) = 1

Therefore, P = 2/3 * 1/2 * 1 = 1/3
:)
This solution is somewhat problematic.

The first part, P(1st letter goes in wrong envelop) = 2/3, is correct.

However, how do we know that P(2nd letter goes in wrong envelop) = 1/2?
The assumption here is that the 1st letter did not go into the envelop intended for the 2nd letter. If the 1st letter did go into the envelop intended for the 2nd letter, then P(2nd letter goes in wrong envelop) = 1 (not 1/2).
This problematic issue also extends to the 3rd letter.
To deal with these issues, we need to consider a lot more cases than what your approach suggests.

Edit: I earlier suggested that this question may be out of scope for the GMAT. However, Mitch's slight adjustment to the solution (below)clears up the complexity issue.

Cheers,
Brent
Thanks Brent for pointing that out.
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