4 letters / 4 envelopes

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yass20015: Senior | Next Rank: 100 Posts; Posts: 38; Joined: Fri Aug 07, 2015 3:06 am; Thanked: 1 times

by yass20015 » Sat Aug 08, 2015 1:47 pm

Tanya prepard 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 into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?
Answer: 1/3. Why ? thanks

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Source: Beat The GMAT — Problem Solving | Sat Aug 08, 2015 1:47 pm

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Aug 08, 2015 1:56 pm

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its complete address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address ?
A. 1/24
B. 1/8
C. 1/4
D. 1/3
E. 3/8

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.

The correct answer is D.

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yass20015: Senior | Next Rank: 100 Posts; Posts: 38; Joined: Fri Aug 07, 2015 3:06 am; Thanked: 1 times

by yass20015 » Sun Aug 09, 2015 3:33 am

Thanks Mitch, but I did not get it. Can you please put it with E1,E2,E3,E4 and L1,L2,L3,L4 ? Many thanks!

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Aug 09, 2015 5:10 am

yass20015 wrote:Thanks Mitch, but I did not get it. Can you please put it with E1,E2,E3,E4 and L1,L2,L3,L4 ? Many thanks!

To my solution above, let's add information about the 4 envelopes.

Let the 4 envelopes be a, b, c and d and the 4 corresponding letters be A, B, C and D.
Let the 4 envelopes be placed on a table in the following order:
a-b-c-d.
Implication:
For every letter to be placed in the correct corresponding envelope, the letters must be arranged in the following order:
ABCD.

Total ways to arrange the 4 letters = 4! = 24.

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.

The correct answer is D.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Aug 09, 2015 6:01 am

Alternate approach:

Let the 4 letters be A, B, C and D.

Case 1: Only A is placed in the correct envelope
P(A is placed in the correct envelope) = 1/4. (Of the 4 envelopes, only 1 corresponds to A.)
P(B is placed in a wrong envelope) = 2/3. (Of the 3 remaining envelopes, 2 are wrong.)
Of the 2 remaining letters, at most 1 can be now be placed in the correct envelope.
P(this letter is placed in the WRONG envelope) = 1/2. (Of the 2 remaining envelopes, 1 is wrong.)
P(last letter is placed in a wrong envelope) = 1/1 = 1. (The 1 remaining envelope must be wrong.)
To combine these probabilities, we multiply:
P(only A is placed in the correct envelope) = 1/4 * 2/3 * 1/2 * 1 = 1/12.

Using the same reasoning:
P(only B is placed in the correct envelope) = 1/12.
P(only C is placed in the correct envelope) = 1/12.
P(only D is placed in the correct envelope) = 1/12.

Since any of the probabilities above will yield a favorable outcome, we ADD the fractions:
1/12 + 1/12 + 1/12 + 1/12 = 4/12 = [spoiler]1/3[/spoiler].

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