Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
30 Mar 2025, 12:55
Kudos
Bookmarks
I solved it by using 8/24 combinatorics method, and it seemed straightforward.
But according to me, I think there's a catch with the commonly used probability method which has been mentioned so many times here.
It resolves to 4*1/4*2/3*1/2*1 and gives the right answer but I believe there's no clear explanation about the third event "Probability of putting third letter in wrong envelope = 1/2" which seems to just correctly work for this question but is not direct.
Let's take an example -
L1 => E1 --- 1/4
L2 => E3/E4 --- 2/3
Case 1 (L2 => E4) =>
L3 => E2 --- 1/2
L4 => E3 --- 1
Works perfectly fine
Case 2 (L2 => E3) =>
L3 => E2/E4 --- 1
Now if we just focus on this event, Letter 3 could have gone to either envelope making the third probability 1 instead of 1/2 as both are incorrect envelopes, but because L4 cannot go to E4 we have to put L3 into E4 making the third probability again 1/2 but this requires us to consider the impact on next event instead of just this individual event.
So this probability technique seems to work in this case, but if there would have been more than 4 envelopes I doubt if the probability of the third event could have been so easily calculated.
@Bunuel/@KarishmaB/@GMATNinja - Can you please help if I am missing anything obvious here?
But according to me, I think there's a catch with the commonly used probability method which has been mentioned so many times here.
It resolves to 4*1/4*2/3*1/2*1 and gives the right answer but I believe there's no clear explanation about the third event "Probability of putting third letter in wrong envelope = 1/2" which seems to just correctly work for this question but is not direct.
Let's take an example -
L1 => E1 --- 1/4
L2 => E3/E4 --- 2/3
Case 1 (L2 => E4) =>
L3 => E2 --- 1/2
L4 => E3 --- 1
Works perfectly fine
Case 2 (L2 => E3) =>
L3 => E2/E4 --- 1
Now if we just focus on this event, Letter 3 could have gone to either envelope making the third probability 1 instead of 1/2 as both are incorrect envelopes, but because L4 cannot go to E4 we have to put L3 into E4 making the third probability again 1/2 but this requires us to consider the impact on next event instead of just this individual event.
So this probability technique seems to work in this case, but if there would have been more than 4 envelopes I doubt if the probability of the third event could have been so easily calculated.
@Bunuel/@KarishmaB/@GMATNinja - Can you please help if I am missing anything obvious here?
25 May 2025, 03:28
Kudos
Bookmarks
Why are we multiplying by 4 here? My understanding- If 3 of 4 are incorrect then possible scenarios- CIII, ICII, IICI or IIIC. Can someone pls confirm if this is the reason or am I missing something?
Wayxi
Wayxi
Quote:
29 Jun 2025, 21:25
Kudos
Bookmarks
Let's imagine four letters L1, L2, L3, L4 and envelopes as E1, E2, E3, E4
L1 --> E1
L2 --> E2
L3 --> E3
L4 --> E4
Total number of arrangements possible are 4!
But we only want one right match i.e. one of (L1, E1), (L2, E2), (L3, E3), (L4, E4)
Above can be repeated 4 times whenever we have different matches like (L2, E2), (L3, E3), (L4, E4)
Total possible = 4 ---------(1)
Let's say you matched L1 with E1 then there is a possibility that others are also matched (like L2 with E2) so we have to exclude them and we can do following in the order:
L2 --> 2 options (like L3 or L4)
L3 --> 1 option (like L4)
L4 --> 1 option (like L2)
Above can be done in 2 ways
Total possible = 2 -----------(2)
Multiplying (1) and (2)
Total number of favorable outcomes = 4*2
Probability = 4*2/(4!) = 1/3
L1 --> E1
L2 --> E2
L3 --> E3
L4 --> E4
Total number of arrangements possible are 4!
But we only want one right match i.e. one of (L1, E1), (L2, E2), (L3, E3), (L4, E4)
Above can be repeated 4 times whenever we have different matches like (L2, E2), (L3, E3), (L4, E4)
Total possible = 4 ---------(1)
Let's say you matched L1 with E1 then there is a possibility that others are also matched (like L2 with E2) so we have to exclude them and we can do following in the order:
L2 --> 2 options (like L3 or L4)
L3 --> 1 option (like L4)
L4 --> 1 option (like L2)
Above can be done in 2 ways
Total possible = 2 -----------(2)
Multiplying (1) and (2)
Total number of favorable outcomes = 4*2
Probability = 4*2/(4!) = 1/3
