difficult prep probability

[mariah]

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter she prepared an envelope with its correct address. If the 4 letters are to be put in 4 envelopes at random, what is the probability

1. that only 1 letter will be put into the envelope with its correct address?

[spoiler]oa 1/3[/spoiler]

2. That no letter will be put into the envelope with its correct address?
[spoiler]oa 9/24[/spoiler]

[mariah]

1.
Thats take we got 4 letters A; B; C: D
and 4 destinations a; b; c; d;

total ways to arrange letters 4! - 24

desired outcomes - 8, thats say if A goes to a, right place, than B C and D have each 2 options
A B C D

a b c d
A has 1 option is goes to a; B has 2 options c & d; C has 2 options b & d ; D - 2 options b and c
hence 1X2X2X2=8 oa [spoiler]8/24[/spoiler]

2. i am stuck with the second problem

it goes same way we got 24 options

now we have to calculate ways each letter goes to wrong way

A B C D

a b c d

A has 3 options b,c , and d; B has 3 options as well a, c, and d ; C and D in same way have each 3 options ..
3X3X3X3=81 - NONSENSE


if it worked in first scenario .. what i am doing wrong, please explain !!!!

[anshumishra]

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter she prepared an envelope with its correct address. If the 4 letters are to be put in 4 envelopes at random, what is the probability

1. that only 1 letter will be put into the envelope with its correct address?

[spoiler]oa 1/3[/spoiler]

2. That no letter will be put into the envelope with its correct address?
[spoiler]oa 9/24[/spoiler]

These are the 24 possible arrangements :

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

1). Probability that only 1 letter will be put into the envelope with its correct address = #arrangements in blue/total# = 8/24 = 1/3

2). Probability that no letter will be put into the envelope with its correct address = #arrangements in brown/total# = 9/24 = 3/8

[Anurag@Gurome]

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter she prepared an envelope with its correct address. If the 4 letters are to be put in 4 envelopes at random, what is the probability

1. that only 1 letter will be put into the envelope with its correct address?

Solution:
Let the envelopes be E1, E2, E3, E4 and the corresponding letters be l1, l2, l3, l4.
Suppose l1 goes to E1 and the other letters do not go in their corresponding envelopes.
So, E2 will have either l3 or l4
If E2 has l3, E3 will have l4, E4 will have l2.
If E2 has l4, E3 will have l2, E4 will have l3.
So, for l1 going to E1 we have 2 arrangements where other letters are not going into right envelopes.
Similarly for l2 going to E2 or l3 going to E3 or l4 going to E4, we have 2 arrangements each.
Or there are 2*2*2*2 = 8 ways in which only one letter goes to right envelope.
Total number of all possible arrangements of letters in any envelopes is 4! = 24.
Or required probability is 8/24 = 1/3.

[Anurag@Gurome]

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter she prepared an envelope with its correct address. If the 4 letters are to be put in 4 envelopes at random, what is the probability


2. That no letter will be put into the envelope with its correct address?

Solution:
Let the envelopes be E1, E2, E3, E4 and the corresponding letters be l1, l2, l3, l4.
If no letter goes into correct address, then let us consider the complementary event.
This will be that either
(a) only 1 letter goes into correct envelope
or (b) only 2 letters go into correct envelopes
or (c) only 3 letters go into correct envelopes
or (d) all 4 letters go into correct envelopes

For (a), we have already seen that number of ways is 8.
Let us consider (b).
If 2 letters are going into right envelopes, there is only 1 way in which other 2 letters are not going into right envelopes.
For example if l1 goes to E1 and l2 goes to E2, l3 will go to E4 and l4 will go to E3.
Now, any 2 letters can be selected from the 4 in 4C2 = 6 ways.
So all 6 pairs will have 6*1 = 6 ways in which only 2 letters go into right envelopes.
Next, consider (c).
This is an impossible situation, since if 3 letters are going right, fourth has to go to the right one.
Lastly, consider (d).
Now, this can happen in only 1 way.
So, the complementary event can happen in 8+6+1 = 15 ways.
Note that 24 is the total number of ways in which any letter can go to any envelope.
So, the number of ways in which no letter goes to correct address is 24 - 15 = 9.
Or, the probability of this occurrence is 9/24 = 3/8.

[mariah]

thanks for your input !

[GMATGuruNY]

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter she prepared an envelope with its correct address. If the 4 letters are to be put in 4 envelopes at random, what is the probability

1. that only 1 letter will be put into the envelope with its correct address?

[spoiler]oa 1/3[/spoiler]

P(exactly n times) = P(one way) * total possible ways.

P(one way):
One way to get a good outcome is for only the first envelope to receive 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, one of the remaining 2 envelopes could still get the correct letter.
P(this envelope gets 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 the correct letter was placed in one of the other envelopes)

Since we need all of these events to happen, we multiply the fractions:
1/4 * 2/3 * 1/2 * 1/1 = 1/12.

Total possible ways:
The result above represents ONE WAY to get a good outcome: if the first envelope is the only one to receive the correct letter.
We need to account for ALL OF THE WAYS to get a good outcome.
Since each of the 4 envelopes could be the one to receive the correct letter, we multiply the result above by 4:
4 * 1/12 = 1/3.

2. That no letter will be put into the envelope with its correct address?
[spoiler]oa 9/24[/spoiler]

The following placements of letters are possible:
None correct
Exactly 1 correct
Exactly 2 correct
All 4 correct (If we have 3 correct, we automatically have all 4 correct.)

P(none) + P(exactly 1) + P(exactly 2) + P(all 4) = 1.
Thus, P(none) = 1 - P(exactly 1) - P(exactly 2) - P(all 4).

P(exactly 1 correct) = 1/3 (as shown above).

P(exactly 2 correct):
P(1st envelope gets the correct letter) = 1/4 (4 letters total, 1 of them correct)
P(next envelope gets the correct letter) = 1/3 (3 letters left, 1 of them correct)
P(next envelope gets 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 the correct letter was placed in the 3rd envelope)

Since we need all of these events to happen, we multiply the fractions:
1/4 * 1/3 * 1/2 * 1/1 = 1/24.

Since, of the 4 envelopes, any combination of 2 could get the correct corresponding letters, the result above needs to be multiplied by 4C2 = 6:
6*24 = 1/4.

P(all 4 correct):
P(1st envelope gets the correct letter) = 1/4 (4 letters total, 1 of them correct)
P(next envelope gets the correct letter) = 1/3 (3 letters left, 1 of them correct)
P(next envelope gets the correct letter) = 1/2 (2 letters left, 1 of them correct)
P(last envelope gets the correct letter) = 1/1 (1 letter left, and it must be correct)

Since we need all of these events to happen, we multiply the fractions:
1/4 * 1/3 * 1/2 * 1/1 = 1/24.

Thus, P(none) = 1 - 1/3 - 1/4 - 1/24 = 1 - 15/24 = 9/24 = 3/8.

[anshumishra]

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter she prepared an envelope with its correct address. If the 4 letters are to be put in 4 envelopes at random, what is the probability

1. that only 1 letter will be put into the envelope with its correct address?

[spoiler]oa 1/3[/spoiler]

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)

P(next envelope gets the wrong letter) = 1/2 (2 letters left, 1 of them wrong)

P(next envelope gets the wrong letter) = 1/1 (1 letter left, and it must be wrong since we placed the correct letter in one of the other envelopes)

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.

Are not we lucky here that the answer matched the OA ?
For example say we want to calculate the probability of first envelope (say A) to be in right envelope :

P(1st envelope, ie. A gets the correct letter) = 1/4 (4 letters total, 1 of them correct), arrangements looks like -> A - - -
P(next envelope, i.e B gets the wrong letter) = 2/3 (3 letters left, 2 of them wrong),arrangements -> AC - - or AD - -
P(next envelope, i.e C gets the wrong letter) = 1 if the arrangement is AC - - or 1/2 if the arrangement is AD --

So, in my opinion you should have branched the probability tree after the two envelopes. Right ?

[GMATGuruNY]

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter she prepared an envelope with its correct address. If the 4 letters are to be put in 4 envelopes at random, what is the probability

1. that only 1 letter will be put into the envelope with its correct address?

[spoiler]oa 1/3[/spoiler]

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)

P(next envelope gets the wrong letter) = 1/2 (2 letters left, 1 of them wrong)

P(next envelope gets the wrong letter) = 1/1 (1 letter left, and it must be wrong since we placed the correct letter in one of the other envelopes)

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.

Are not we lucky here that the answer matched the OA ?
For example say we want to calculate the probability of first envelope (say A) to be in right envelope :

P(1st envelope, ie. A gets the correct letter) = 1/4 (4 letters total, 1 of them correct), arrangements looks like -> A - - -
P(next envelope, i.e B gets the wrong letter) = 2/3 (3 letters left, 2 of them wrong),arrangements -> AC - - or AD - -
P(next envelope, i.e C gets the wrong letter) = 1 if the arrangement is AC - - or 1/2 if the arrangement is AD --

So, in my opinion you should have branched the probability tree after the two envelopes. Right ?

Once we're down to the last 2 envelopes, we have 2 letters left.
One of these 2 remaining envelopes could still get the correct corresponding letter.
Let C = the envelope that could still get the correct corresponding letter.
P(C gets the wrong letter) = 1/2.
This would force the last envelope to get the wrong letter.
Thus, P(last envelope gets the wrong letter) = 1.

I've amended my original post so that the reasoning is clearer.

[anshumishra]

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter she prepared an envelope with its correct address. If the 4 letters are to be put in 4 envelopes at random, what is the probability

1. that only 1 letter will be put into the envelope with its correct address?

[spoiler]oa 1/3[/spoiler]

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)

P(next envelope gets the wrong letter) = 1/2 (2 letters left, 1 of them wrong)

P(next envelope gets the wrong letter) = 1/1 (1 letter left, and it must be wrong since we placed the correct letter in one of the other envelopes)

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.

Are not we lucky here that the answer matched the OA ?
For example say we want to calculate the probability of first envelope (say A) to be in right envelope :

P(1st envelope, ie. A gets the correct letter) = 1/4 (4 letters total, 1 of them correct), arrangements looks like -> A - - -
P(next envelope, i.e B gets the wrong letter) = 2/3 (3 letters left, 2 of them wrong),arrangements -> AC - - or AD - -
P(next envelope, i.e C gets the wrong letter) = 1 if the arrangement is AC - - or 1/2 if the arrangement is AD --

So, in my opinion you should have branched the probability tree after the two envelopes. Right ?

Once we're down to the last 2 envelopes, we have 2 letters left.
One of these 2 remaining envelopes could still get the correct corresponding letter.
Let C = the envelope that could still get the correct corresponding letter.
P(C gets the wrong letter) = 1/2.
This would force the last envelope to get the wrong letter.
Thus, P(last envelope gets the wrong letter) = 1.

I've amended my original post so that the reasoning is clearer.

Yes, the new reasoning is justified. The second solution had already these things taken into account, so no need to change anything there.

[OneTwoThreeFour]

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter she prepared an envelope with its correct address. If the 4 letters are to be put in 4 envelopes at random, what is the probability

1. that only 1 letter will be put into the envelope with its correct address?

[spoiler]oa 1/3[/spoiler]

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, one of the remaining 2 envelopes could still get the correct letter.
P(this envelope gets 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 the correct letter was placed in one of the other envelopes)

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.
[spoiler]oa 9/24[/spoiler]

Why aren't we taking into consideration of the different ways that any 1 letter could be put into the correct envelope? For example, we have letters L1, L2, L3, L4, and envelopes E1, E2, E3, E4. If we put L1 into E1 on the first try, and the rest of the letters are in the wrong envelopes, then the probability is 1/4*2/3*1/2*1 = 1/12.

But lets say we put L2 into E3 on the first try, and L1 into E1 on the second try, and the rest of the letters are put into their respective envelopes incorrectly, shouldn't the probability be:

3/4 * 1/3* 1/2*1= 1/8. (I-C-I-I) (Let "I" stands for incorrect and "C" stands for correct.)

If we put L1 into E1 on the third try and the rest of the letters are put into their envelopes incorrectly:
3/4 * 2/3 * 1/2 *1 = 1/4 (I-I-C-I)

If we put L1 into E1 on the fourth try and the rest of the letters are put into their envelopes incorrectly:
3/4 * 2/3 * 1/2*1=1/4 (I-I-I-C)

So shouldn't it be 1/12 + 1/8 + 1/4 + 1/4= 17/24.

Since we have four different letters, the final probability should be 4 * 17/24, or 68/24. This is definately wrong since any probability cannot be greater than 1, so why is my reasoning wrong?

P(exactly 2 correct):
P(1st envelope gets the correct letter) = 1/4 (4 letters total, 1 of them correct)
P(next envelope gets the correct letter) = 1/3 (3 letters left, 1 of them correct)
P(next envelope gets 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 the correct letter was placed in the 3rd envelope)

Since we need all of these events to happen, we multiply the fractions:
1/4 * 1/3 * 1/2 * 1/1 = 1/24.

Since, of the 4 envelopes, any combination of 2 could get the correct corresponding letters, the result above needs to be multiplied by 4C2 = 6:
6*24 = 1/4.

When calculating two letters will be put into the correct envelope, we are taking into consideration of the different ordering of C-C-I-I. But when we are calculating the probably that exactly one letter will be put into the correct envelope, why aren't we taking into consideration the different ordering of C-I-I-I?

Thanks!

[/quote]

[saketk]

Guys -- question 2 is nothing but a scenario of Derangements.. (Google it if you have not studied this in school)

the forumla is

D(n) = n! - n!/1! + n!/2! - n!/3! + n!/4!+....(-1)^n*n!/n!

Using this forumla we can calculate the probability --

D(4) = 4!-4!+4!/2!-4!/3!+4!/4!
= 13-4
= 9

Thus our answer is
9/24 [where 24 is the total number of arrangements]