Problem Solving

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 four letters to be put into envelopes at random, what is the probability that only one letter will put in to the envelope with ots correct address?

a) 1/8 b) 1/24 c) 3/8 d)1/3 e)1/2

Probability of Only A put into correct = 1/4 * 2/3 * 1/2 * 1

Similarly for other three as well, the probabilty will be same

Total probability = 4 * (1/4 * 2/3 * 1/2 * 1) = 1/3

Answer [spoiler]{D}[/spoiler]??

akpareek 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 four letters to be put into envelopes at random, what is the probability that only one letter will put in to the envelope with ots correct address?

a) 1/8 b) 1/24 c) 3/8 d)1/3 e)1/2

Approach 1:

Let the 4 letters be A, B, C and D.
Total ways to arrange the 4 letters = 4! = 24.
Let the correct ordering of the 4 letters be ABCD.

Write out the ways that ONLY A can be put in the correct position:
ACDB
ADBC
Total ways = 2.

Using the same reasoning, there will be 2 ways that ONLY B can be put in the correct position, 2 ways that ONLY C can be put in the correct position, and 2 ways that ONLY D can be put in the correct position.
Thus, the total number of ways to put EXACTLY 1 letter in the correct position = 2+2+2+2 = 8.

Thus:
P(exactly 1 letter is put in the correct position) = 8/24 = 1/3.

Approach 2:

Let the correct order for the letters be A-B-C-D.

Case 1: Only A in the correct envelope
P(A is placed in the CORRECT envelope) = 1/4. (Of the 4 envelopes, 1 is correct.)
P(B is placed in an INCORRECT envelope) = 2/3. (Of the 3 remaining envelopes, only 1 is correct, so 2 are incorrect.)

At this point, B has taken either the envelope for C or the envelope for D.
Thus, of C and D, ONLY ONE could now be placed in the correct envelope.
P(this letter is placed in an INCORRECT envelope) = 1/2. (Of the 2 remaining envelopes, only 1 is correct, so 1 is incorrect.)
P(last letter is placed in an INCORRECT envelope) = 1. (As noted above, only ONE of C and D could be placed in the correct envelope, so the other must be placed in an incorrect envelope.)

Since all of these events must happen in order for A to be the only envelope correctly placed, we multiply the fractions:
1/4 * 2/3 * 1/2 * 1.

Other cases:
Since there are 4 options for the one envelope correctly placed -- A, B, C, or D -- the result above must be multiplied by 4:
1/4 * 2/3 * 1/2 * 1 * 4 = 1/3.

Approach 3:

A DERANGEMENT is a permutation in which NO element is in the correct position.
Given n elements (where n>1):

Number of derangements = n! (1/2! - 1/3! + 1/4! + ... + ((-1)^n)/n!)

Let the 4 letters be ABCD.

Total arrangements:
Total number of ways to arrange the 4 letters = 4! = 24.

Good arrangements:
In a good arrangement, EXACTLY ONE letter is in the correct position.
Number of options for the one letter put into the correct position = 4. (A, B, C, or D)
Number of ways to DERANGE the 3 remaining letters = 3! (1/2! - 1/3!) = 3-1 = 2.
To combine these options, we multiply:
4*2 = 8.

Good/total = 8/24 = 1/3.

The correct answer is D.

Envelopes A,B,C,D have correct letter in each:

P (only A) = p(A) * p(not B) * p(not C) * p(not D)

P (only A) = 1/4 * 2/3 * 1/2 * 1 = 1/12

Same for only B, C or D

So P(only A, B, C or D) = 4 * 1/12 = 1/3