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HARD PROB

by nycknicks11 » Tue Jun 07, 2011 10:10 pm
Tanya prepared four different letters to be sent to four different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?

1/24
1/8
1/4
1/3
3/8

My approach: 4 slots _ _ _ _: they can be wither R or W. So there are 4 ways to be right and 2^4 total combination. Answer: 4/16= 1/4.


After I did this problem out by brute force, I understand why my method is wrong but can someone explain to me intuitively why that method is wrong. I want to truly understand the "signal" for me to not use that method again in other problems.
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by cans » Tue Jun 07, 2011 10:17 pm
let letters be A,B,C,D
total ways of putting 4 letters in 4 envelopes = 4! (take one letter and you have 4 options to put it into. Take 2nd letter, and you have only 3 options left and so on..)
only one letter goes into its correct address.
No. of ways of selecting that particular letter = 4C1 = 4.
Suppose we select D.
Now A,B,C should all go in wrong envelopes. Let correct envelopes for A,B,C be P,Q,R respectively.
A can't go in P. Thus only 2 options (either Q or R)
Suppose it goes into Q. Then C can't go into R and Q is already occupied. Thus Q goes into P. (only 1 option). B goes into R (that's the only one remaining)
Thus ways = 4*2*1*1 = 8
prob = 8/4! = 1/3
IMO D
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by nycknicks11 » Tue Jun 07, 2011 10:33 pm
good stuff. Do you know why my method of using binary (R or W) for 4 slots is wrong intuitively? I use the binary method all the time. Kinda scary that it does not work sometimes.

(Here's the reason why it doesn't work. This is not intuitive though. 2^4 represents ALL combination of R or W arranged in 4 spots. HOWEVER, you cannot have RRRW. Doesn't work. If the other 3 is right the last is also right. So you have to subtract all combinations of 3 R and 1 W. So 4/(2^16-4)= 1/3).
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by phanideepak » Tue Jun 07, 2011 10:41 pm
As you said there are 4 places to fill so naturally since the order matters the number of arrangements are 4!

and also lets calculate the probability for the letter to be right only in the first position so for the second position there are only 2 possible options to chose from as 1 is the right letter.

So for the third position there is only one option to chose from

and for the fourth position we are left with only one number so no choices.

so the probability that the letter goes right only in the first place is given by 2/4!

Now the probablity that the letter goes only into one place is 4C1 X(2/4!) so answer is

[spoiler]1/3[/spoiler]
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by phanideepak » Tue Jun 07, 2011 10:43 pm
nycknicks11 wrote:good stuff. Do you know why my method of using binary (R or W) for 4 slots is wrong intuitively? I use the binary method all the time. Kinda scary that it does not work sometimes.

(Here's the reason why it doesn't work. This is not intuitive though. 2^4 represents ALL combination of R or W arranged in 4 spots. HOWEVER, you cannot have RRRW. Doesn't work. If the other 3 is right the last is also right. So you have to subtract all combinations of 3 R and 1 W. So 4/(2^16-4)= 1/3).
Right you would have been right provided the events would have been independent of each other :) But here we have to consider the domino effect as in if you get one letter in the wrong place you are also messing up another envelope in this case :)
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by Frankenstein » Tue Jun 07, 2011 10:51 pm
nycknicks11 wrote:Tanya prepared four different letters to be sent to four different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?

1/24
1/8
1/4
1/3
3/8

My approach: 4 slots _ _ _ _: they can be wither R or W. So there are 4 ways to be right and 2^4 total combination. Answer: 4/16= 1/4.


After I did this problem out by brute force, I understand why my method is wrong but can someone explain to me intuitively why that method is wrong. I want to truly understand the "signal" for me to not use that method again in other problems.
Hi,
Cans has already explained the correct answer. I will try to explain the flaw in your method.
1)Firstly, in a slot R can be in only 1 way but W can be in 3 ways.
So, consider R,W,W,W. According to you this is only 1 iteration. But, in fact it can be achieved in 2 ways.
Example Let the right sequence be 1,2,3,4. Your condition of R,W,W,W can be achieved in 2 ways:
1,3,4,2
1,4,2,3.
So you have missed 4 patterns in this way for 1R and 3Ws
2)Similarly you have missed 8 more patterns for 4Ws
Example: W,W,W,W can be achieved in 9 ways:
3,4,2,1
3,1,4,2
3,4,1,2
2,4,1,3
2,3,4,1
2,1,4,3
4,1,2,3
4,3,2,1
4,3,1,2
3)You are also considering the possibility of 3Rs and 1W, which is impossible because 3Rs makes the 4th one right.
So you have considered 4 patterns which are invalid
So your denominator will be 16+4+8-4 = 24.
As, mentioned above 1R,3Ws case will occur in 8 ways instead of 4 ways.
So, probability is 8/24 = 1/3.
Last edited by Frankenstein on Tue Jun 07, 2011 11:06 pm, edited 1 time in total.
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by nycknicks11 » Tue Jun 07, 2011 11:05 pm
you ever use the binary method? 70% of the time on prob. problem I prefer to use combination. 90% of those times you have to reduce it to a binary problem.

Frankenstein wrote:
nycknicks11 wrote:Tanya prepared four different letters to be sent to four different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?

1/24
1/8
1/4
1/3
3/8

My approach: 4 slots _ _ _ _: they can be wither R or W. So there are 4 ways to be right and 2^4 total combination. Answer: 4/16= 1/4.


After I did this problem out by brute force, I understand why my method is wrong but can someone explain to me intuitively why that method is wrong. I want to truly understand the "signal" for me to not use that method again in other problems.
Hi,
Cans has already explained the correct answer. I will try to explain the flaw in your method.
Firstly, in a slot R can be in only 1 way but W can be in 3 ways.
So, consider R,W,W,W. According to you this is only 1 iteration. But, in fact it can be achieved in 2 ways.
Example Let the right sequence be 1,2,3,4. Your condition of R,W,W,W can be achieved in 2 ways:
1,3,4,2
1,4,2,3.
So you have missed 4 patterns in this way for 1R and 3Ws
Similarly you have missed 4 more patterns for 4Ws
Example: W,W,W,W can be achieved in 8 ways:
3,4,2,1
3,1,4,2
3,4,1,2
2,4,1,3
2,3,4,1
2,1,4,3
4,1,2,3
4,3,2,1
4,3,1,2
So your denominator will be 16+4+4 = 24.
As, mentioned above 1R,3Ws case will occur in 8 ways instead of 4 ways.
So, probability is 8/24 = 1/3.

Sorry, I have missed something here.. Will edit it in few minutes.
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by Frankenstein » Tue Jun 07, 2011 11:09 pm
nycknicks11 wrote:you ever use the binary method? 70% of the time on prob. problem I prefer to use combination. 90% of those times you have to reduce it to a binary problem.
Hi,
Yes, As Phanideepak has rightly pointed, binary method is applicable only when the events are independent. My lengthy post was just to show you the cases you haven't considered and the invalid ones you considered.
Example:
1)Number of 4 digits numbers formed from a set of {1,2,3,4} if digits can be repeated(Independent).
2)Number of 4 digits numbers formed from a set of {1,2,3,4} if digits cannot be repeated.

Lastly, kindly post such questions in the Problem Solving section(not DS section).
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by GMATGuruNY » Wed Jun 08, 2011 2:15 am
nycknicks11 wrote:Tanya prepared four different letters to be sent to four different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?

1/24
1/8
1/4
1/3
3/8

My approach: 4 slots _ _ _ _: they can be wither R or W. So there are 4 ways to be right and 2^4 total combination. Answer: 4/16= 1/4.


After I did this problem out by brute force, I understand why my method is wrong but can someone explain to me intuitively why that method is wrong. I want to truly understand the "signal" for me to not use that method again in other problems.
P(only the 1st envelope gets the correct letter):
P(1st envelope gets the correct letter) = 1/4 (4 letters total, 1 of them correct)
P(next envelope gets the wrong letter) = 2/3 (3 letters left, 2 of them wrong)

At this point, 1 of the 2 remaining envelopes could still get the correct letter.
P(this envelope the wrong letter) = 1/2 (2 letters left, 1 of them wrong)
P(last envelope gets the wrong letter) = 1/1 (1 letter left, and it must be wrong)

Since we need all of these events to happen, we multiply the fractions:

1/4 * 2/3 * 1/2 * 1/1 = 1/12.

Since each envelope has the same probability of getting the correct letter and we have 4 envelopes total, we need to multiply by 4:

4 * 1/12 = 1/3.

The correct answer is D.
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by bubbliiiiiiii » Wed Jun 08, 2011 2:45 am
Hi GMATGuruNY,

Could you please elaborate this? I am not clear with this statement.
Since each envelope has the same probability of getting the correct letter and we have 4 envelopes total, we need to multiply by 4:

4 * 1/12 = 1/3.
Hi GMATGuruNY,

Could you please elaborate this? I am not clear with this statement.
Regards,

Pranay
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by GMATGuruNY » Wed Jun 08, 2011 3:07 am
bubbliiiiiiii wrote:Hi GMATGuruNY,

Could you please elaborate this? I am not clear with this statement.
Since each envelope has the same probability of getting the correct letter and we have 4 envelopes total, we need to multiply by 4:

4 * 1/12 = 1/3.
Hi GMATGuruNY,

Could you please elaborate this? I am not clear with this statement.
Let's call the 4 envelopes E1, E2, E3 and E4.

The probability that only E1 (the "1st envelope" in my solution above) gets the correct letter = 1/12.

But E1 is not the only envelope that could get the correct letter. Each of the 4 envelopes has the same probability of getting the correct letter:

P(E1 gets the correct letter) = 1/12.
P(E2 gets the correct letter) = 1/12.
P(E3 gets the correct letter) = 1/12.
P(E4 gets the correct letter) = 1/12.

Thus, the probability that exactly one of the envelopes gets the correct letter = 1/12 + 1/12 + 1/12 + 1/12 = 4 * 1/12 = 1/3.
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by magicalhat » Wed Jun 08, 2011 8:37 am
yes, i'm confused, too.
bubbliiiiiiii wrote:Hi GMATGuruNY,

Could you please elaborate this? I am not clear with this statement.
Since each envelope has the same probability of getting the correct letter and we have 4 envelopes total, we need to multiply by 4:

4 * 1/12 = 1/3.
Hi GMATGuruNY,

Could you please elaborate this? I am not clear with this statement.
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by bubbliiiiiiii » Wed Jun 08, 2011 8:46 pm
Thanks Mitch for coming back on this topic and explaining the statement.

@Magicalhat,

Till the statement quoted above we have found the probability of letter 1 being into envelope 1 and all other letters i.e., 2,3 and 4 are placed in wrong envelopes. Hope you agree with that.

Now why do we multiply by 4?

Here it is, letter 2 can also be placed in envelope 2 while all other letters i.e., 1, 3 and 4 can be placed in wrong envelopes. If we calculate the probability of doing this, it will be same as that of first case.

Similary, for letter 3 and letter 4.

So in total we have 4 cases by now which have the same probability.

In such cases we always add the four probabilities. Since the probabilities of all the four cases are same in this case we can either add all four of them or directly multiply the probability of first case with 4 to include all the four cases.

Hope it makes sense. :)
Regards,

Pranay
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by gmat1978 » Sun Jun 12, 2011 10:50 pm
Cans,

Good explanation. Can you please explain the following?
cans wrote: No. of ways of selecting that particular letter = 4C1 = 4.
Suppose we select D.
I got it wrong. A,B,C,D are 4 letters and P,Q,R,S are corresponding envelopes. Let's put A in the correct envelope(P). We can do that only in 1 way right? Can you please point out the flaw in my thought process?

Thanks
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by Frankenstein » Sun Jun 12, 2011 10:54 pm
gmat1978 wrote:
cans wrote: No. of ways of selecting that particular letter = 4C1 = 4.
Suppose we select D.
I got it wrong. A,B,C,D are 4 letters and P,Q,R,S are corresponding envelopes. Let's put A in the correct envelope(P). We can do that only in 1 way right? Can you please point out the flaw in my thought process?

Thanks
Hi,
Yes.. If you pick A, you will put it in correct envelope in 1 way. Similar in another case, you will B and put in correct envelope Q. Similarly for the other 2 letters. So, By adding all these, it gives you 4 ways.
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