GPrep 2

[mclaren7]

Hi Ron

You the man. Much appreciated.

KH

[TanyaHater]

The question asks specifically for the times when exactly 1 letter is put into the envelope with its correct address. When Ron listed his 24 possibilities, he put asterisks next to the ones that matched this criterion - there were 8 of these. So there are 8 desired possibilities out of 24 total outcomes.

If you're confused about why he asterisked those 8, I recommend writing the problem out for yourself the way he did it, but doing what he suggested: arrange the letters vertically so that you can see more easily which ones match your desired scenario)

Stacy, I still do not understand, how Ron got the 8 desired scenarios.

[RonPurewal]

The question asks specifically for the times when exactly 1 letter is put into the envelope with its correct address. When Ron listed his 24 possibilities, he put asterisks next to the ones that matched this criterion - there were 8 of these. So there are 8 desired possibilities out of 24 total outcomes.

If you're confused about why he asterisked those 8, I recommend writing the problem out for yourself the way he did it, but doing what he suggested: arrange the letters vertically so that you can see more easily which ones match your desired scenario)

Stacy, I still do not understand, how Ron got the 8 desired scenarios.

you're looking for scenarios in which exactly one letter is placed in the correct envelope.

if you write out groups of 4 letters, corresponding to which letters go in envelopes a., b., c., and d., then this means that EXACTLY ONE of the following 4 statements is true:
* 'a' is in the first position
* 'b' is in the second position
* 'c' is in the third position
* 'd' is in the fourth position.
the eight situations with asterisks are the eight situations in which exactly one of these four things is true.

[shaji]

Hi friends,

My above post in summary:

1. The probability that ONLY A matches
The probability of Letter A going into Envelope A = 1/4
The probability of Letter B NOT going into Envelope B = 2/3
The probability of Letter C NOT going into Envelope C = 1/2
The probability of Letter D NOT going into Envelope D = 1
ONLY A matches is 1/4 x 2/3 x 1/2 x 1 = 2/24 = 1/12

Since there are 4 sets = 4 x 1/12 = 1/3.

HOWEVER, there is a slight problem if we proceed on via addition of probability.

2. The probability that ONLY B matches
The probability of Letter A NOT going into Envelope A = 3/4
The probability of Letter B going into Envelope B = 1/3
The probability of Letter C NOT going into Envelope C = 1/2
The probability of Letter D NOT going into Envelope D = 1
ONLY B matches 3/4 x 1/3 x 1/2 x 1 = 1/8

3. The probability that ONLY C matches
The probability of Letter A NOT going into Envelope A = 3/4
The probability of Letter B NOT going into Envelope B = 2/3
The probability of Letter C going into Envelope C = 1/2
The probability of Letter D NOT going into Envelope D = 1
ONLY C matches 3/4 x 2/3 x 1/2 x 1 = 1/4

4. The probability that ONLY D matches
The probability of Letter A NOT going into Envelope A = 3/4
The probability of Letter B NOT going into Envelope B = 2/3
The probability of Letter C NOT going into Envelope C = 1/2
The probability of Letter D going into Envelope D = 1
ONLY D matches 3/4 x 2/3 x 1/2 x 1 = 1/4

Probability is 17/24.
Sigh. Where did I go wrong?
For multiplication of probabilities, it seems fine, but you add up the individual probabilites, it doesn't add up.

KH

[shaji]

you have to be extremely careful whenever you multiply consecutive probabilities for statements that encompass multiple possibilities (especially negative statements, like the ones here). let me point out what can go wrong, using your first group of calculations as an example:

Hi friends,

My above post in summary:

1. The probability that ONLY A matches
The probability of Letter A going into Envelope A = 1/4
The probability of Letter B NOT going into Envelope B = 2/3
The probability of Letter C NOT going into Envelope C = 1/2
The probability of Letter D NOT going into Envelope D = 1
ONLY A matches is 1/4 x 2/3 x 1/2 x 1 = 2/24 = 1/12

Since there are 4 sets = 4 x 1/12 = 1/3.

there are issues with this calculation. it happens to hit upon the correct value at the end, but that's a total coincidence.

i agree with the first two probabilities: the probability that letter a goes into envelope a is indeed 1/4, and the probability if that happens that the letter b goes into an envelope other than b is 2/3.
however, it's downhill from there: if letter b actually went into envelope c, then the probability of letter c not going into envelope c is 1. the probability is only 1/2 (as you've stated) if letter b winds up in envelope d.
similarly, the final probability is either 0 or 1, depending on whether the last envelope remaining is envelope d or not. if letter b goes in envelope c and letter c goes in envelope b (fulfilling all of your conditions), then letter d is stuck going into envelope d, making that last probability 0.

so, if you're going to go this route, you're stuck with doing the following:
* first 2 steps = same as you have them now
* 3rd step = 2 branches of a probability tree, depending on whether envelope c is still available (vs. whether it was used for letter b)
* 4th step = 2 branches off EACH of those prior 2 branches, depending on whether envelope d is still available (vs. whether it was used for letter b or c)

that really, really sucks.

in general, if you have a COMBINATION problem with a SMALL # OF COMBINATIONS, you are much better off just counting possibilities than trying to employ fancy probability tricks.

hth!

A very clear example of compensating errors. Some mathematicans term them 'good mistakes' or more simply getting the correct answer for the wrong reasons.

[rfernandez]

Glad it was helpful.

[Guest]

Hi instructors,

Would it be correct to solve this problem in this manner as well- 1*2*1*1 + 2*1*1*1+ 2*1*1*1 + 2*1*1*1 / 4! = 1/3
The probability of getting the correct letter in the correct envelope is 1 the first time around, then the probability of getting the wrong envelope in the second pick is 2 since there are 3 remaining envelopes and so on... and then just reverse the order since we have for letters to deal with.

Your input is greatly appreciated...

[RonPurewal]

Hi instructors,

Would it be correct to solve this problem in this manner as well- 1*2*1*1 + 2*1*1*1+ 2*1*1*1 + 2*1*1*1 / 4! = 1/3
The probability of getting the correct letter in the correct envelope is 1 the first time around, then the probability of getting the wrong envelope in the second pick is 2 since there are 3 remaining envelopes and so on... and then just reverse the order since we have for letters to deal with.

Your input is greatly appreciated...

no, invalid solution. the fact that this yields the correct number is pure happenstance.
this approach is already debunked in one of my earlier posts in this thread, dated 27 Feb 2008 04:54 am. check it out.

[eatGMAT4breakfast]

This post might be too late for those who asked the questions but for others who are still preparing:

Think of the envelope 1-A, 2-B, 3-C, 4-D (relationship between the correct envelope and address)

1234
ABCD (all four to the correct address)
ABDC (2 match)
ACBD (2 match)
ACDB (only 1 matching ... this is what we want)
ADBC (only 1)
ADCB (2 matching)

Now you can least the scenarious starting with B, C than D. We have 2x4 = 8 winners.

Hope it helps anyone who is currently preparing.

Cheers!

[RonPurewal]

This post might be too late for those who asked the questions but for others who are still preparing:

Think of the envelope 1-A, 2-B, 3-C, 4-D (relationship between the correct envelope and address)

1234
ABCD (all four to the correct address)
ABDC (2 match)
ACBD (2 match)
ACDB (only 1 matching ... this is what we want)
ADBC (only 1)
ADCB (2 matching)

Now you can least the scenarious starting with B, C than D. We have 2x4 = 8 winners.

Hope it helps anyone who is currently preparing.

Cheers!

invalid reasoning.

the fact that this works is, again, a complete coincidence. in fact, you should be suspicious of this method right out of the gate: all of the instances you've listed have letter A in envelope A. those are obviously "special", and you shouldn't expect the rest of the instances to behave similarly.

for instance, extending your reasoning, since none of the examples in your list have 0 matches, you would conclude that 4 x 0 = NONE of the overall possibilities have 0 matches.
this is, unfortunately, wrong, as there are plenty of ways to have 0 matches (such as bcda, cdab, dcba, etc. - lots of them).

in short:
don't assume symmetry unless your examples are actually symmetric.
here, they aren't. you just get really stratospherically lucky.

i am, however, astonished at the number of COMPLETELY WRONG approaches that "accidentally" get the right answer to this problem. given the difficulty level of the problem and the unusually high amount of labor involved in solving it, i can't help but wonder whether it was chosen that way on purpose.

[prude_sb]

One simple way of looking at this

calculate total number of possibilities

first envelope has 4 possible letters that can go in, second envelope has 3 and so on

so 4! is the total number

now select one envelope for which you are going to put in the right letter - 4 ways

we have 3 envelopes remaining ... for the first envelope you have 2 possible letter that can go in (one of them is the correct letter so it is ruled out) - 2 ways

there is only one way of putting the remaining two letters in the two envelopes such that they are put inside the wrong envelopes - 1 way

so total for one letter to be in the right envelope = 4 * 2 * 1

so probability is 8/4! = 1/3

[Kweku.Amoako]

Here is another way of doing this problems

There 4 letters and 4 addresses

Lets pick the first letter and figure out the probability of putting in the right envelope

1) One right address and 3 wrong addresses so the probability is (1/4)

Next lets figure our how to place the next letter into a wrong address: One evelope has already been picked out in (1) so we have 3 envelopes left.

(2) Of of these 3, 2 have the wrong address so the probability of picking the wrong address is (2/3)

Next lets figure out the probability of picking a wrong address for the 3rd letter

(3) 2 addresses with one wrong address so probability is (1/2)

(4) Probability of picking the wrong address for the 4th is 1


so that probability is (1/4)(2/3)(1/2) ...remember this for the first letter. We can repeat the process for all 4 letters or just multiply this expression by 4

4(1/4)(2/3)(1/2) = 1/3

[RonPurewal]

the approach in the previous post is flawed.
it gets the correct answer, by pure coincidence, but it doesn't work. let me point out the problem.

(by the way, this exact approach has been debunked earlier in this thread, but the thread is so long that it's easier for me just to type again what's wrong with it.)

Here is another way of doing this problems

There 4 letters and 4 addresses

Lets pick the first letter and figure out the probability of putting in the right envelope

1) One right address and 3 wrong addresses so the probability is (1/4)

Next lets figure our how to place the next letter into a wrong address: One evelope has already been picked out in (1) so we have 3 envelopes left.

(2) Of of these 3, 2 have the wrong address so the probability of picking the wrong address is (2/3)

Next lets figure out the probability of picking a wrong address for the 3rd letter

(3) 2 addresses with one wrong address so probability is (1/2)

that last calculation is problematic, because there may not be "one wrong address" among the two remaining ones.
for instance, let's say that you first put letter A into envelope A.
now let's say you're dealing with letter B (whose probability you've given as 2/3 for the wrong envelope), and THEN letter C (whose probability you've given as 1/3 for the wrong envelope).
if you put letter B into envelope D, then you'll indeed have a probability of 1/2 for letter C, since the remaining envelopes will be B and C.
if you put letter B into envelope C, on the other hand, you'll have a probability of 1 for letter C, since the two remaining envelopes (B and D) are both wrong.

there are other problems with this approach, too, notably the implication that the FIRST letter goes into the correct envelope. (this approach neglects possibilities in which the second, third, or fourth letter goes into the correct envelope.)

--

still, this approach, despite its flaws, gives the correct answer, so luck wins out here.

in any case, this is an extremely difficult problem, and would no doubt be highly relevant only to the scores of people scoring 750+ on the exam.

[tigerwoods]

WHERE IS THE QUESTION I DONT SEE IT ?? IN THE POST

[RonPurewal]

WHERE IS THE QUESTION I DONT SEE IT ?? IN THE POST

there are two pages of posts. (if anyone posts after this, there will be three pages.) you have to go to the beginning of the first page to see the original post.

you can either click "previous page" or click "page 1"; the page-number links can be found both at the top and at the bottom of the thread.