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30 Mar 2025, 12:55
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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
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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
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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
