GPrep 2

[Nov1907]

Could someone help with a quick solution for this question? Thanks!

[shaji]

Could someone help with a quick solution for this question? Thanks!

Interesting Problem with practical conotations.

The four possibilities are 0,1,2 and 4 in the correct envelope/s.

The possibilities of these curring are: 9,8,6,1

The prob of 1 letter in the right evelope is therfore 1/3.

[RonPurewal]

Hey people - common-sense guideline: if you're going to post anything, then post the EXPLANATIONS for your answers. If you just post some numbers and say 'these are the possibilities,' then no one is really being helped.

This is one of those problems on which, if you're stuck, you can just make a list and count things: the total number of possibilities is only 4! = 24. You can list that many possibilities in well under a minute, especially if you arrange them in an organized manner.

Here we go: Use ALPHABETICAL ORDER.

ABCD ABDC ACBD ACDB* ADBC* ADCB
BACD BADC BCAD* BCDA BDAC BDCA*
CABD* CADB CBAD CBDA* CDAB CDBA
DABC DACB* DBAC* DBCA DCAB DCBA

I've marked with asterisks those arrangements in which exactly one letter goes in the matching envelope (this is a lot easier to see if you arrange the letters vertically, with the heading ABCD at the top, as you might do on the yellow dry-erase pad).

What you should get from this discussion: If you're stuck on a problem, don't be afraid to just list things and do the problem the 'brute-force way'. Really, don't. Every second you spend just staring at a problem is a second wasted.

[RonPurewal]

I should add that, because there are 8 starred combinations above, the probability in question is 8 / 24 = 1/3.

[shaji]

The 'brute force way' proves the distibution of possibilities and is indeed a practical approach. If Tanya had more than 4 letters and envelopes, things will turn very 'brutal' this way.

Now the math involved is a bit difficullt to explain on this form, as it involves Venn Diagrams, Baye's theorem(Conditional Probability) and combinotrics. Appling these will enable you to solve it in a single line and would not take even 45 secs.

in this prob, it goes like this:

Prob(only1 in correct envelope)=(1/2+1/6-1/24)-(1/2+1/6+1/24)=1/3

Hey people - common-sense guideline: if you're going to post anything, then post the EXPLANATIONS for your answers. If you just post some numbers and say 'these are the possibilities,' then no one is really being helped.

This is one of those problems on which, if you're stuck, you can just make a list and count things: the total number of possibilities is only 4! = 24. You can list that many possibilities in well under a minute, especially if you arrange them in an organized manner.

Here we go: Use ALPHABETICAL ORDER.

ABCD ABDC ACBD ACDB* ADBC* ADCB
BACD BADC BCAD* BCDA BDAC BDCA*
CABD* CADB CBAD CBDA* CDAB CDBA
DABC DACB* DBAC* DBCA DCAB DCBA

I've marked with asterisks those arrangements in which exactly one letter goes in the matching envelope (this is a lot easier to see if you arrange the letters vertically, with the heading ABCD at the top, as you might do on the yellow dry-erase pad).

What you should get from this discussion: If you're stuck on a problem, don't be afraid to just list things and do the problem the 'brute-force way'. Really, don't. Every second you spend just staring at a problem is a second wasted.

[shaji]

Soory , i bracket missed with a -sign.
Prob(only1 in correct envelope)=(1/2+1/6-1/24)-(1/2-(1/6+1/24))=1/3

The 'brute force way' proves the distibution of possibilities and is indeed a practical approach. If Tanya had more than 4 letters and envelopes, things will turn very 'brutal' this way.

Now the math involved is a bit difficullt to explain on this form, as it involves Venn Diagrams, Baye's theorem(Conditional Probability) and combinotrics. Appling these will enable you to solve it in a single line and would not take even 45 secs.

in this prob, it goes like this:

Prob(only1 in correct envelope)=(1/2+1/6-1/24)-(1/2+1/6+1/24)=1/3

Hey people - common-sense guideline: if you're going to post anything, then post the EXPLANATIONS for your answers. If you just post some numbers and say 'these are the possibilities,' then no one is really being helped.

This is one of those problems on which, if you're stuck, you can just make a list and count things: the total number of possibilities is only 4! = 24. You can list that many possibilities in well under a minute, especially if you arrange them in an organized manner.

Here we go: Use ALPHABETICAL ORDER.

ABCD ABDC ACBD ACDB* ADBC* ADCB
BACD BADC BCAD* BCDA BDAC BDCA*
CABD* CADB CBAD CBDA* CDAB CDBA
DABC DACB* DBAC* DBCA DCAB DCBA

I've marked with asterisks those arrangements in which exactly one letter goes in the matching envelope (this is a lot easier to see if you arrange the letters vertically, with the heading ABCD at the top, as you might do on the yellow dry-erase pad).

What you should get from this discussion: If you're stuck on a problem, don't be afraid to just list things and do the problem the 'brute-force way'. Really, don't. Every second you spend just staring at a problem is a second wasted.

[Guest]

Hey people - common-sense guideline: if you're going to post anything, then post the EXPLANATIONS for your answers. If you just post some numbers and say 'these are the possibilities,' then no one is really being helped.

This is one of those problems on which, if you're stuck, you can just make a list and count things: the total number of possibilities is only 4! = 24. You can list that many possibilities in well under a minute, especially if you arrange them in an organized manner.

Here we go: Use ALPHABETICAL ORDER.

ABCD ABDC ACBD ACDB* ADBC* ADCB
BACD BADC BCAD* BCDA BDAC BDCA*
CABD* CADB CBAD CBDA* CDAB CDBA
DABC DACB* DBAC* DBCA DCAB DCBA

I've marked with asterisks those arrangements in which exactly one letter goes in the matching envelope (this is a lot easier to see if you arrange the letters vertically, with the heading ABCD at the top, as you might do on the yellow dry-erase pad).

What you should get from this discussion: If you're stuck on a problem, don't be afraid to just list things and do the problem the 'brute-force way'. Really, don't. Every second you spend just staring at a problem is a second wasted.

Hi Ron,

I get the 24 possibilities for this problem. Can you explain how you get 8 winning scenarios?

Thanks!

[StaceyKoprince]

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)

[vietst]

where i am wrong when I do like this? W: wrong, R: right
No right: WWWW: 1 case - 4C0
1 right: RWWW, WRWW, WWRW and WWWR- 4 cases - 4C1
2 right: RRWW, RWRW, RWWR, WWRR, WRWR, WRRW - 6 cases - 4C2
3 right: RRRW, RRWR, RWRR, WRRR - 4 cases - 4C3
4 right: RRRR - 1 case - 4C4
Total: 16 cases
1 right: 4 cases
then 4/16 = 1/4

How we can expand the case of 5 letters?

[StaceyKoprince]

There are more than 16 cases b/c some of the orderings that you set up have multiple possible set-ups.

For example, in this case:
First, let's use A, B, C, D to represent the four letters (and, if they are in the right envelopes, they'll be in that order - A in the first slot, B in the second slot, and so on)

For your RWWW, that could be ACDB, for example. Or it could be ADBC. Both of those have the first one (A) in the right envelope and all of the other three in the wrong envelopes - but they represent two different scenarios, not just the one generic scenario of RWWW. (Note that you couldn't do, say, ACBD, because now D is in the right slot as well - so there's a limit to the number of ways you can arrange the three "wrong" letters.)

By contrast, for your RRRR, that would have to be ABCD - there are no other options there, so there's really only one RRRR scenario.

So, in your set-up, you've counted some of the possible arrangements, but not all of them.

[vietst]

Stacey Koprince,
I have known this. Thanks

[RonPurewal]

There are more than 16 cases b/c some of the orderings that you set up have multiple possible set-ups.

For example, in this case:
First, let's use A, B, C, D to represent the four letters (and, if they are in the right envelopes, they'll be in that order - A in the first slot, B in the second slot, and so on)

For your RWWW, that could be ACDB, for example. Or it could be ADBC. Both of those have the first one (A) in the right envelope and all of the other three in the wrong envelopes - but they represent two different scenarios, not just the one generic scenario of RWWW. (Note that you couldn't do, say, ACBD, because now D is in the right slot as well - so there's a limit to the number of ways you can arrange the three "wrong" letters.)

By contrast, for your RRRR, that would have to be ABCD - there are no other options there, so there's really only one RRRR scenario.

So, in your set-up, you've counted some of the possible arrangements, but not all of them.

you bet-cha.

...not to mention the fact that all of the possibilities with exactly three r's (rrrw, rrwr, rwrr, wrrr) are just plain impossible. if you don't see why, think about stuffing the first 3 envelopes correctly, and then think about the only possible thing that could happen with the fourth envelope.

[mclaren7]

Dear friends,

Add my 2 cents' worth, would it be correct to do the following way?

For the 1st letter to be placed with the correct address, and the rest of the 3 sets being mismatched, probability is:
= 1/4 [probability of one letter being corrected matched with its envelope] x 2/3 [P of 2nd letter being WRONGLY matched with the envelope] x 1/2 [P of 3rd letter being wrongly matched with its envelope] x 1 [last letter with its wrong envelope]
= 1/12.

Since there are 4 letters, we need to multiply x 4 = 4/12 = 1/3.
C = correct set, W = wrong set
CWWW
WCWW
WWCW
WWWC

Correct?

KH

[mclaren7]

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

[RonPurewal]

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!