Q) Stacy prepared 4 different letters to be sent to 4 addresses. For each letter she prepared an envelope with its correct address. If the 4 letters to be put in to 4 envelopes at random, what is the probability that only one letter will be put in to the envelope with the correct address?

How do we solve this...

Thanks,
Champ

There are several ways to solve this.
Notation: 1R3W means 1 Right, 3 Wrong

Method 1: The "# of ways method" or "the combinatorics method"
Probability = # of ways to get 1R3W/# of ways total

# 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.

# of ways to get 1R3W is more complicated. 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.

Probability is 8/24 = 1/3.

Method 2: The successive probability method.
We’ll call the letters A, B, C, and D, and make lists of ways the letters could be placed in envelopes, with the assumption that A would correctly go in the first envelope, B belongs in the second, etc.

There are 4! = 24 total cases. We'll look first at RWWW probability (i.e. A is the only letter in the correct envelope.)

1/4 of the ways to assign letters are of the form Axxx: ABCD, ABDC, ACBD, ACDB, ADBC, ADCB
2/3 of those have a “wrong” letter in the second envelope (i.e. not B): ACBD, ACDB, ADBC, ADCB
3/4 of these have a “wrong” letter in the third envelope (i.e. not C): ACBD, ACDB, ADBC
2/3 of these have a “wrong” letter in the forth envelope (i.e. not D): ACDB, ADBC
Combined, there is a (1/4)(2/3)(3/4)(2/3) = 1/12 chance of RWWW.

As noted above, the RWWW chance is the same as the chance of WRWW or WWRW or WWWR, so the total chance of 1R3W is 4*1/12 = 1/3.

_________________
Emily Sledge
Instructor
ManhattanGMAT

Thanks, Emily.

You are welcome. My thanks go to Ron for catching a logical error in Method 2, now fixed.

_________________
Emily Sledge
Instructor
ManhattanGMAT

I have a question on the above:

[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.]

When you go RWWW - you get 1*2*1*1

But say you try to calculate the options for WRWW.

If you fill the first envelope with the wrong letter (you have 3 wrong letters to choose from), then fill the 2nd one with the right letter (you now don't know if the *right* letter for this one was chosen as the first *wrong letter - so you don't know how many different options there are) and so on.....you can't get an answer.

Am I missing something, or is the general guidance to ALWAYS take the constraint as the first choice, and then multiply it by no. of variations??

Thanks!

The number of ways available to distribute the letters doesn't depend on which envelope you stuff first. There's no relevant difference among the envelopes that makes one more likely than another to be the only envelope to receive the correct letter, so start with the simplest account, RWWW, then multiply by 4.

In case you find that frustrating--Why doesn't the order matter?--consider the options for arriving at WRWW. Let r1= the letter right for envelope 1, etc. The two possible orderings are r3r2r4r1 and r4r2r1r3.

Probability is stated as the ratio of the number of favourable cases to the total number of cases.
Here's a formula which u guys might find useful.
Here i am stating the formula in the context of a problem.
Suppose 'n' persons go to a bar with each having an umbrella with a mark on it so that each can recognise his/her own umbrealla.
After getting drunk,they began leaving the bar one by one.
If the question asks to find the number of ways in which none of them leaves the bar with their own umbrealla then how it is to be done.(here order of people leaving the bar is irrelevant.Only thing important is that each person leaves with an umbrella not his/her own)
The formula is this: n!*(1-1/1!+1/2!-1/3!+.......+((-1)^n)*1/n!)(n=the number of persons)
The above formula gives the number of ways in which the people leave the bar with a different umbrella.
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For the above problem..
The favourable case where one letter goes into the right envelope is 4C1(=4) and the number of ways in which the rest three letters go into the wrong envelopes is 3!*(1-1/1!+1/2!-1/3!)(=2).
Hence required probablity in which one letter goes into the right envelope and the rest three letters go into the wrong envelopes is 4*2=8.

Now the number of ways in which the letters can go into the envelopes(with no restriction)=4!=24.

Hence the required answer is 8/24=1/3.
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I have tried my best to clarify this formula.In case u feel it difficult to understand,i am ready to try again.But this formula is really awesome for these kinda problems.

Hey Guys,

I am able to find the total number of favourable outcomes. However, there is some problem that I encountered while taking the total number of outcomes.

The method Emily told seems good to me 4*3*2*1 = 24.

However, it will be great if someone can explain what I am doing wrong in what I am doing to calculate the total number of outcome.

There are 4 letters and there are 4 envelops.
There are 4c1 ways of selecting a letter and 4c1 for selecting the envelops. So total number of ways = 4*4 = 16.

Not sure if I am missing something, please explain.
Thanks.

that's just the number of ways of selecting ONE letter and ONE envelope.

that's not what you have to do in this problem; in this problem, you must place EACH of the letters into a different envelope. therefore, a single outcome must specify the envelope in which each of the four letters, not just one of them, is placed.