Problem Solving

Q. 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?

One of the answer explanations which came was -

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
My Doubt
let us say the letters and envelopes are L1,E1,L2,E2,L3,E3,L4,E4

now let us take 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 picks E3, then prob of incorrect envelope for L3 will be 1 (only E2 and E4 left) and similarly prob of incorrect envelope for L4 will be 1 too...

am i thinking correct??

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.

Another approach, assume there are 4 empty spaces (envelops) that have to be filled by one letter each - A, B, C and D. Total number of ways to do this = 4*3*2*1 = 24 ways

Now if we assume that the correct line up is ABCD. Further assume, first space is filled correctly in how many ways can we do that 1 way (A goes there), 2nd pace has to be filled wrong how many ways = 2 (fill by C or D) 3rd place has to be filled wrong how many ways to do that (B or C OR B or D depending on what was picked earlier) and finally the 4th place is by default wrong.
Thus total number of ways = 2. Now we can fix B and rotate the others similarly C and D can be fixed hence there are 4 ways to do this, hence total number of favorable cases = 4*2 = 8

Hence favorable probability = 8/24 =1/3

gotcha...thanks much

mj78ind wrote:Another approach, assume there are 4 empty spaces (envelops) that have to be filled by one letter each - A, B, C and D. Total number of ways to do this = 4*3*2*1 = 24 ways

Now if we assume that the correct line up is ABCD. Further assume, first space is filled correctly in how many ways can we do that 1 way (A goes there), 2nd pace has to be filled wrong how many ways = 2 (fill by C or D) 3rd place has to be filled wrong how many ways to do that (B or C OR B or D depending on what was picked earlier) and finally the 4th place is by default wrong.
Thus total number of ways = 2. Now we can fix B and rotate the others similarly C and D can be fixed hence there are 4 ways to do this, hence total number of favorable cases = 4*2 = 8

Hence favorable probability = 8/24 =1/3