Probability

gmatmillenium wrote:Tanya prepared 4 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 into 4 envelopes at random, what is the probability that only 1 letter will be put its envelope with correct address?

C=correct
I=incorrect
so lets say the order of letters is CIII
now first letter can be put in correct envelope by 1/4
now second letter can be put in incorrect envelope by 2/3
now third letter can be put in incorrect envelope by 1/2
now fourth letter can be put in incorrect envelope by 1
1/4+2/3+1/2+1=1/12

now CIII can be rearranged in 4C1=4 ways

probability=4*1/12=1/3 way

btw this Q should be moved to problem solving forum and not CR

Hi Outreach

Thanks for replying and my apologies for not acknowledging earlier...

here is my related doubt...let us say the letters and envelopes are L1,E1,L2,E2,L3,E3,L4,E4

now let us take your CIII....

Prob of correct envelope for L1 = 1/4
Prob of incorrect envelope for L2 = 2/3....(here L2 can pick E3 or E4)
If L2 goes into E3, then prob of incorrect envelope for L3 will be 1 and similarly prob of incorrect envelope for L4 will be 1 too...

can you please explain this?

outreach wrote:

gmatmillenium wrote:Tanya prepared 4 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 into 4 envelopes at random, what is the probability that only 1 letter will be put its envelope with correct address?

C=correct
I=incorrect
so lets say the order of letters is CIII
now first letter can be put in correct envelope by 1/4
now second letter can be put in incorrect envelope by 2/3
now third letter can be put in incorrect envelope by 1/2
now fourth letter can be put in incorrect envelope by 1
1/4+2/3+1/2+1=1/12

now CIII can be rearranged in 4C1=4 ways

probability=4*1/12=1/3 way

deleted because I think there was something wrong with my logic!

gmatmillenium wrote:Tanya prepared 4 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 into 4 envelopes at random, what is the probability that only 1 letter will be put its envelope with correct address?

Probability is defined as:

(total number of good outcomes)/(total number of possible outcomes)

Sometimes the best approach is to determine each part of the fraction separately.

Let's say the 4 letters are ABCD, and to place each in the correct envelope, they have to be placed in alphabetical order: ABCD.

To determine the total number of possible outcomes, we need to determine how many ways we can arrange the 4 letters:

4 * 3 * 2 * 1 = 24 total possible outcomes.

A good outcome is when 1 letter is in the correct position and the other 3 are not.

Let's say we want only letter A to be in the right position. We'd have only 2 good arrangements:

ACDB and ADBC

2 good outcomes so far.

Extending this logic, we know that:
If we want only B to be in the right position, we'd get another 2 good outcomes.
If we want only C to be in the right position, we'd get another 2 good outcomes.
If we want only D to be in the right position, we'd get another 2 good outcomes.

So altogether we have 2 + 2 + 2 + 2 = 8 good outcomes.

So (good outcomes)/(total outcomes) = 8/24 = 1/3.

Hi Mitch,

How did you arrive at "2" good outcomes with every letter? why only ACDB and ADBC for A?....this logic is perhaps my missing link

Thanks

GMATGuruNY wrote:

gmatmillenium wrote:Tanya prepared 4 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 into 4 envelopes at random, what is the probability that only 1 letter will be put its envelope with correct address?

Probability is defined as:

(total number of good outcomes)/(total number of possible outcomes)

Sometimes the best approach is to determine each part of the fraction separately.

Let's say the 4 letters are ABCD, and to place each in the correct envelope, they have to be placed in alphabetical order: ABCD.

To determine the total number of possible outcomes, we need to determine how many ways we can arrange the 4 letters:

4 * 3 * 2 * 1 = 24 total possible outcomes.

A good outcome is when 1 letter is in the correct position and the other 3 are not.

Let's say we want only letter A to be in the right position. We'd have only 2 good arrangements:

ACDB and ADBC

2 good outcomes so far.

Extending this logic, we know that:
If we want only B to be in the right position, we'd get another 2 good outcomes.
If we want only C to be in the right position, we'd get another 2 good outcomes.
If we want only D to be in the right position, we'd get another 2 good outcomes.

So altogether we have 2 + 2 + 2 + 2 = 8 good outcomes.

So (good outcomes)/(total outcomes) = 8/24 = 1/3.

gmatmillenium wrote:Hi Mitch,

How did you arrive at "2" good outcomes with every letter? why only ACDB and ADBC for A?....this logic is perhaps my missing link

Thanks

If we correctly put A in the 1st position, we have BCD left.

We can't put B in the 2nd position, so our options for 2nd position are C and D, giving us AC or AD in the first 2 positions.

If we start with AC, we have to finish with DB, because we can't put D in the last position, giving us: ACDB

If we start with AD, we have to finish with BC, because we can't put C in the 3rd position, giving us ADBC.

So if A is in the 1st postion, we have two good outcomes: ACDB and ADBC.

Make sense?

Thanks much Mitch.....

GMATGuruNY wrote:

gmatmillenium wrote:Hi Mitch,

How did you arrive at "2" good outcomes with every letter? why only ACDB and ADBC for A?....this logic is perhaps my missing link

Thanks

If we correctly put A in the 1st position, we have BCD left.

We can't put B in the 2nd position, so our options for 2nd position are C and D, giving us AC or AD in the first 2 positions.

If we start with AC, we have to finish with DB, because we can't put D in the last position, giving us: ACDB

If we start with AD, we have to finish with BC, because we can't put C in the 3rd position, giving us ADBC.

So if A is in the 1st postion, we have two good outcomes: ACDB and ADBC.

Make sense?