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Probability of 1 of 4 letters reaching right addressed env

Expert replies
by utkalnayak » Fri Jan 23, 2015 6:14 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

With all these letters and envelopes things together I lost my way !!
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Source: — Problem Solving |

by utkalnayak » Fri Jan 23, 2015 6:15 pm
OA : [spoiler]1/3[/spoiler]
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by GMATGuruNY » Fri Jan 23, 2015 6:41 pm
utkalnayak wrote: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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by utkalnayak » Fri Jan 23, 2015 6:47 pm
GMATGuruNY wrote: 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.
In the above case why you think there are only 2 possibilities in each case ?
Should it not be 3! ways ?
Thanks,
Utkal
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by [email protected] » Fri Jan 23, 2015 10:03 pm
Hi utkalnayak,

In this prompt, you have 2 conditions that have to be met:

1) JUST 1 letter is put in the correct envelope
2) The other 3 letters must be put in an INCORRECT envelope.

IF...."A" is properly placed in the "A" envelope, B CANNOT be put in the "B" envelope, C CANNOT be put in the "C" envelope and D cannot be put in the "D" envelope. Those other rules still have to be accounted for.

Mitch's solution correctly accounts for BOTH conditions.

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by Matt@VeritasPrep » Sun Jan 25, 2015 3:30 pm
utkalnayak wrote:In the above case why you think there are only 2 possibilities in each case ?
Should it not be 3! ways ?
Here's a visual that might help.

Image
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