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Re: Tanya prepared 4 different letters to 4 different addresses [#permalink]
31 Aug 2013, 05:35

Expert's post

bagdbmba wrote:

Hi Guys, Can anybody shot an explanation to my query raised above ?

I'd much appreciate if any Quant expert kindly explain this issue to me!

This case is solved in the following way: Counting all incorrect, or 0 correct: P(all incorrect)=1-(1 correct)-(2 correct)-(3 correct)-(4 correct)=1-8/24-6/24-0-1/24=9/24.

Don;t know what I can add to that... _________________

Re: Tanya prepared 4 different letters to 4 different addresses [#permalink]
15 Sep 2013, 11:23

1

This post received KUDOS

We are trying to find the probability of 1R3W.

Probability = number of ways to get 1R3W/number of ways total

number of ways total is 4! = 24. Imagine stuffing envelopes randomly. Stacy can put any of 4 letters into the first envelope, any of the remaining 3 into the next, either of the remaining 2 into the next, and has no choice to make on the last, or 4*3*2*1.

number of ways to get 1R3W : She could fill the first envelope with the right letter (1 way), then put either of the 2 wrong remaining letters in the next (2 ways), then put a wrong letter in the next (1 way). That's 1*2*1*1 = 2.

But since it doesn't have to be the first envelope that has the Right letter, it could be any of the 4 envelopes (i.e. we could have RWWW, WRWW, WWRW, WWWR), the total ways to get 1R3W is 4*2 = 8.

Re: Tanya prepared 4 different letters to 4 different addresses [#permalink]
21 May 2014, 04:35

Aximili85 wrote:

GODSPEED wrote:

My 2 cents:

For 1 letter to go in right envelope - probability = 1/4 For 2nd letter to go in wrong envelope - probability = 2/3 For 3rd letter to go in wrong envelope - probability = 1/2 For 4th letter to go in wrong envelope - probability = 1

Now, in the order this arrangement can be done: C - Letter going into correct envelope W - Letter going into wrong envelope

No#1. C - W - W - W ............in correct envelope: 4* (1/4) * (2/3) * (1/2) * 1 = 1/3

Thats the only explanation i was able to understand =( I'm worried because if I don't learn the combinatrics formula way of doing it I might make a logical inconsistency while solving under a time limit.

Just wondering, is the above method correct? What if the correct is not the first one? Does it change the procedure?

Re: Tanya prepared 4 different letters to 4 different addresses [#permalink]
21 May 2014, 04:35

Aximili85 wrote:

GODSPEED wrote:

My 2 cents:

For 1 letter to go in right envelope - probability = 1/4 For 2nd letter to go in wrong envelope - probability = 2/3 For 3rd letter to go in wrong envelope - probability = 1/2 For 4th letter to go in wrong envelope - probability = 1

Now, in the order this arrangement can be done: C - Letter going into correct envelope W - Letter going into wrong envelope

No#1. C - W - W - W ............in correct envelope: 4* (1/4) * (2/3) * (1/2) * 1 = 1/3

Thats the only explanation i was able to understand =( I'm worried because if I don't learn the combinatrics formula way of doing it I might make a logical inconsistency while solving under a time limit.

Just wondering, is the above method correct? What if the correct is not the first one? Does it change the procedure?

Tanya prepared 4 different letters to 4 different addresses [#permalink]
20 Sep 2014, 16:21

jlgdr wrote:

Just wondering, is the above method correct? What if the correct is not the first one? Does it change the procedure?

Please advice Cheers! J

I actually had the same question and was about to post a reply on this thread seeking same but while typing it had the aha moment.

Yes, even when one calculates the probability of each scenario separately, the total will be \(\frac{1}{3}\). #1 - C W W W - \(\frac{1}{4}*\frac{2}{3}*\frac{1}{2}*1=\frac{1}{12}\) #2 - W C W W - \(\frac{2}{4}*\frac{1}{3}*\frac{1}{2}*1=\frac{1}{12}\) (Note for the first W, desired outcomes are only two, the 3rd and 4th envelopes, and doesn't include its own envelope and the envelope for the second letter as that needs to have the correct one. Same explanation plays in scenarios #3 and #4) #3 - W W C W - \(\frac{2}{4}*\frac{1}{3}*\frac{1}{2}*1=\frac{1}{12}\) #4 - W W W C - \(\frac{2}{4}*\frac{1}{3}*\frac{1}{2}*1=\frac{1}{12}\)

Re: Tanya prepared 4 different letters to 4 different addresses [#permalink]
01 Dec 2014, 02:09

Dienekes wrote:

jlgdr wrote:

Just wondering, is the above method correct? What if the correct is not the first one? Does it change the procedure?

Please advice Cheers! J

I actually had the same question and was about to post a reply on this thread seeking same but while typing it had the aha moment.

Yes, even when one calculates the probability of each scenario separately, the total will be \(\frac{1}{3}\). #1 - C W W W - \(\frac{1}{4}*\frac{2}{3}*\frac{1}{2}*1=\frac{1}{12}\) #2 - W C W W - \(\frac{2}{4}*\frac{1}{3}*\frac{1}{2}*1=\frac{1}{12}\) (Note for the first W, desired outcomes are only two, the 3rd and 4th envelopes, and doesn't include its own envelope and the envelope for the second letter as that needs to have the correct one. Same explanation plays in scenarios #3 and #4) #3 - W W C W - \(\frac{2}{4}*\frac{1}{3}*\frac{1}{2}*1=\frac{1}{12}\) #4 - W W W C - \(\frac{2}{4}*\frac{1}{3}*\frac{1}{2}*1=\frac{1}{12}\)

Bunuel, can you check this post, pls. I think there are mistakes in calculations of 2nd, 3rd, and 4th conditions. For the first W I think there must 3/4. Am I right? _________________

If my post was helpful, press Kudos. If not, then just press Kudos !!!

Re: Tanya prepared 4 different letters to 4 different addresses [#permalink]
14 Mar 2015, 09:53

Bunuel wrote:

dk94588 wrote:

Hello, this was on GMATprep, and I have had problems with this type of question before, but maybe you could help me solve it.

Tanya prepared 4 different letters to 4 different addresses. For each letter, she prepared one envelope with its correct address. If the 4 letters are to be put into the four envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?

A) 1/24 B) 1/8 C) 1/4 D) 1/3 E) 3/8

Total # of ways of assigning 4 letters to 4 envelopes is \(4!=24\).

Only one letter in the right envelope: 4(# of envelopes)*2(# of ways possible to arrange 3 letters incorrectly in the envelopes, when one is correct).

ABCD(envelopes) ACDB(letters) ADBC(letters) (When A is in the right envelope other three have only 2 possible incorrect arrangements) As we have 4 letters, total # of ways 4*2=8

Hi Bunuel, what if there were 5 letters and we were asked to find the same probability as in the original question , how would we calculate these possible wrong arrangements using combitronics ? (When A is in the right envelope other three have only 2 possible incorrect arrangements) ABCDE Total arrangements for BCDE=4! case 1: only 2 wrong : 4 case 2: three wrong : X so all wrong will be 4!-4 -X ?

could you please help .

thanks lucky _________________

Thanks, Lucky

_______________________________________________________ Kindly press the to appreciate my post !!

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