BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Featured Course
Target Test Prep GMAT OnDemand
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
- 715+ score guarantee — highest in the industry (99th percentile)
- 52 chapters · 1,500+ lessons · 4,000+ practice questions
- 400 hours of video · 1,500+ instructor-led HD smartboard lessons
- 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
- 24/7 live support + weekly Zoom office hours with GMAT instructors
- TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
- 1,200+ flashcards + AI-powered study assistant & daily calendar
- OnDemand, LiveTeach & GMAT Bootcamp formats available
- Also: GRE, SAT Math & Executive Assessment courses
- MBA Admissions Consulting now available
- 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
- 200,000+ students served
- 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★
5.0
(559 reviews)
130-pt guarantee
$0 to start
then $127/mo
GMAT Prep2 (Probability)Envelopes.
Source: Beat The GMAT — Problem Solving |
I just thought of a better way:
Probability of putting first letter in correct envelope: 1/4
Probability of putting second letter in wrong envelope: 2/3
Probability of putting third letter in wrong envelope: 1/2
Probability of putting last letter in wrong envelope: 1/1
Multiply all those together and you get 1/12. We have four envelopes, so (1/12)*4 = 1/3.
Probability of putting first letter in correct envelope: 1/4
Probability of putting second letter in wrong envelope: 2/3
Probability of putting third letter in wrong envelope: 1/2
Probability of putting last letter in wrong envelope: 1/1
Multiply all those together and you get 1/12. We have four envelopes, so (1/12)*4 = 1/3.
Jim S. | GMAT Instructor | Veritas Prep
Hi jim,
The last part is not clear. why have u multiplied with 4. You have already taken into account 4 envelops and putting letters in 3 of them wrongly !
The last part is not clear. why have u multiplied with 4. You have already taken into account 4 envelops and putting letters in 3 of them wrongly !
Such kind of problems are called 'Derangements' in probability
Total Number of ways of arranging 4 letters in 4 envelopes = 4!
First one letter is selected in 4C1 way and put in the correct envelope.
Now we have to see that the remaining 3 letters are put in the wrong envelopes which is given by (3!) [1/2! - 1/3! ]
Thus Probability of only one letter being put in correct envelope
= 4C1 * (3!) [1/2! - 1/3! ] / 4! = 1/3
Neo
Total Number of ways of arranging 4 letters in 4 envelopes = 4!
First one letter is selected in 4C1 way and put in the correct envelope.
Now we have to see that the remaining 3 letters are put in the wrong envelopes which is given by (3!) [1/2! - 1/3! ]
Thus Probability of only one letter being put in correct envelope
= 4C1 * (3!) [1/2! - 1/3! ] / 4! = 1/3
Neo
MBA : My passion and My pursuit
ksh- I think that's because there are 4 different ways that that the "winning" scenario that Jim explained can occur (4 envelopes)ksh wrote:Hi jim,
The last part is not clear. why have u multiplied with 4. You have already taken into account 4 envelops and putting letters in 3 of them wrongly !
Hi getneonow
Im sorry but i could not understand your solution completely.
To be precise this part:
:
[i]Now we have to see that the remaining 3 letters are put in the wrong envelopes which is given by (3!) [1/2! - 1/3! ] [/i]
--------------------
This is how i am trying to solve:
Letter 1 => Can be any one of the four letters , i.e. 4 ways to pick a letter - and 1 correct way to place that one into the right envelope.
Therefore, for the first letter => 4*1 ways to place in correct envelope
For the remaining 3 letter:
3 ways to pick aletter => 2 ways to place in incorrect envelope ==> 3*2
2 ways to pick a letter => 1 ways to place in incorrect envelope ==> 2*1
1 ways to pick a letter => 2 ways to place in incorrect envelope ==> 1*1
Total number of ways for (1 correct & 3 incorrect) = (4*1) * (3*2) * (2*1) * (1*1) = 48 ways
Total ways to place 4 letters in 4 envelopes => (4*4) * (3*3) * (2*2) * (1*1) = 576 ways
Probability (1 correct & 3 incorrect) = 48/576 = 1/12
----------------------------------------------------------------
I would appreciate if someone can explain how is this wrong.
And how to reach the correct solution.
Thanking in advance.
Im sorry but i could not understand your solution completely.
To be precise this part:
: [i]Now we have to see that the remaining 3 letters are put in the wrong envelopes which is given by (3!) [1/2! - 1/3! ] [/i]
--------------------
This is how i am trying to solve:
Letter 1 => Can be any one of the four letters , i.e. 4 ways to pick a letter - and 1 correct way to place that one into the right envelope.
Therefore, for the first letter => 4*1 ways to place in correct envelope
For the remaining 3 letter:
3 ways to pick aletter => 2 ways to place in incorrect envelope ==> 3*2
2 ways to pick a letter => 1 ways to place in incorrect envelope ==> 2*1
1 ways to pick a letter => 2 ways to place in incorrect envelope ==> 1*1
Total number of ways for (1 correct & 3 incorrect) = (4*1) * (3*2) * (2*1) * (1*1) = 48 ways
Total ways to place 4 letters in 4 envelopes => (4*4) * (3*3) * (2*2) * (1*1) = 576 ways
Probability (1 correct & 3 incorrect) = 48/576 = 1/12
----------------------------------------------------------------
I would appreciate if someone can explain how is this wrong.
And how to reach the correct solution.
Thanking in advance.
onesome..I have used a generic formula!! If u are unaware of it, better not follow that approach.onesome wrote:Hi getneonow
Im sorry but i could not understand your solution completely.
To be precise this part:
Now we have to see that the remaining 3 letters are put in the wrong envelopes which is given by (3!) [1/2! - 1/3! ]
--------------------
This is how i am trying to solve:
Letter 1 => Can be any one of the four letters , i.e. 4 ways to pick a letter - and 1 correct way to place that one into the right envelope.
Therefore, for the first letter => 4*1 ways to place in correct envelope
For the remaining 3 letter:
3 ways to pick aletter => 2 ways to place in incorrect envelope ==> 3*2
2 ways to pick a letter => 1 ways to place in incorrect envelope ==> 2*1
1 ways to pick a letter => 2 ways to place in incorrect envelope ==> 1*1
Total number of ways for (1 correct & 3 incorrect) = (4*1) * (3*2) * (2*1) * (1*1) = 48 ways
Total ways to place 4 letters in 4 envelopes => (4*4) * (3*3) * (2*2) * (1*1) = 576 ways
Probability (1 correct & 3 incorrect) = 48/576 = 1/12
----------------------------------------------------------------
I would appreciate if someone can explain how is this wrong.
And how to reach the correct solution.
Thanking in advance.
There are other logical ways as VP_Jim did.
Even yours is a good approach but has a subtle wrong elements in it.
You are unnecessarily calculating the task of PICKING of each letter, which is totally not required.
We have been asked to put only one correct letter in correct envelope. So its sufficient if you are calculating the number of ways of picking that letter and putting it in right envelope which you have done it right.
Letter 1 => Can be any one of the four letters , i.e. 4 ways to pick a letter - and 1 correct way to place that one into the right envelope.
Therefore, for the first letter => 4*1 ways to place in correct envelope
But after this step, it shouldn't really matter in how many ways you are picking up the letters to go in wrong envelopes. Simply try to calculate if one of the remaining letters is taken then what are the number of ways for it to go to wrong envelope.
i.e, For letter 2 -> 2 ways
For letter 3 -> 1 way
For letter 4-> 1 way
So total 4*2*1*1 = 8 ways
and Number of ways of arranging 4 letters in 4 envelopes is simply 4! but not (4!)^2 . Here also you are repeating the same mistake of PICKING up the letters.
When you say Letter1, it is specific and you just want to bother how you would place that letter in 4 envelopes which is given by 4 ways only
Similarly for others it is in 3,2,1 ways..
totally 4! ways
so probability = 8/4! = 1/3
Hope this helps
Neo
MBA : My passion and My pursuit
Could you please explain the approach if i make following change in the question.
The letters are marked A, B, C and D, Find out the probability the A goes in the right enevelope.
The letters are marked A, B, C and D, Find out the probability the A goes in the right enevelope.
Note that you have already marked the letters here and specifically asking the first letter A to go to correct envelope.durgesh79 wrote:Could you please explain the approach if i make following change in the question.
The letters are marked A, B, C and D, Find out the probability the A goes in the right enevelope.
# of ways 4 letters can be put in 4 envelopes = 4!
# of ways A can be put in correct envelope = 1
# of ways B is put in wrong envelope = 2
# of ways C is put in wrong envelope = 1
# of ways C is put in wrong envelope = 1
Total # of ways = 1*2*1*1 = 2
Prob = 2/4! = 1/12
Neo
MBA : My passion and My pursuit
Thanks Neo,
Makes perfect sense.
Now the probability that only one of A, B, C or D goes into the right envelope will be 1/12+1/12+1/12+1/12 = 1/3
Makes perfect sense.
Now the probability that only one of A, B, C or D goes into the right envelope will be 1/12+1/12+1/12+1/12 = 1/3
Bingo!!durgesh79 wrote:Thanks Neo,
Makes perfect sense.
Now the probability that only one of A, B, C or D goes into the right envelope will be 1/12+1/12+1/12+1/12 = 1/3
Or rather its simply 4 * 1/12 (for 4 letters) = 1/3
Neo
MBA : My passion and My pursuit
Thanks getneonow.. I truly appreciate it 
Thanks to durgesh79 as well - the way it was elaborated in following 2 posts was just perfect.
Thank you guys.
Thanks to durgesh79 as well - the way it was elaborated in following 2 posts was just perfect.
Thank you guys.
