GPrep 2 :: Manhattan GMAT Forums

Posted: Wed Feb 27, 2008 9:56 am

Hi Ron

You the man. Much appreciated.

KH

Posted: Wed May 14, 2008 6:17 pm

skoprince wrote:

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.

Posted: Mon May 19, 2008 4:56 am

TanyaHater wrote:

skoprince wrote:

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.

Posted: Mon May 19, 2008 6:15 am

mclaren7 wrote:

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

Posted: Mon May 19, 2008 6:20 am

RPurewal wrote:

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:

mclaren7 wrote:

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.

Posted: Fri Jun 06, 2008 3:30 am

Glad it was helpful.

Posted: Thu Aug 14, 2008 6:38 pm

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...

Posted: Mon Aug 25, 2008 2:23 am

Anonymous wrote:

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.