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by prachi18oct » Sun Aug 31, 2014 9:54 am
Image


I used below approach:-
Total favorable outcomes = 4-> RWWW,WRWW,WWRW,WWWR where W and R stand for Wrong address and Right address envelope.
Total outcomes = 4!
Probability of only one correct = 4/24 = 1/6 ?
This is not even in the options.I have been banging my head for hours but dont understand how my logic is flawed. Experts please help!
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by GMATGuruNY » Sun Aug 31, 2014 10:16 am
prachi18oct wrote: I used below approach:-
Total favorable outcomes = 4-> RWWW,WRWW,WWRW,WWWR where W and R stand for Wrong address and Right address envelope.
Total outcomes = 4!
Probability of only one correct = 4/24 = 1/6 ?
This is not even in the options.I have been banging my head for hours but dont understand how my logic is flawed. Experts please help!
The portion in red undercounts the number of favorable outcomes.
Tanya prepared 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?

(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 GMATinsight » Tue Sep 02, 2014 8:49 pm
prachi18oct wrote:Image

Total Outcomes of putting 4 letters into 4 envelopes randomly = 4 x 3 x 2 x 1 = 4! = 24

Favorable cases = 4 (selecting the letter that goes into correct envelope let' say A) x 2 (second letter to go into wrong envelope let's say B goes into C's envelope) x 1(third letter to go into wrong envelope i.e. C will have to go into D's envelope which if not done the D will have to go into right envelope) x 1(D goes into B's envelope)

i.e. Favorable cases = 4 x 2 x 1 x 1 = 8

Favorable Probability = 8/24 = 1/3

Answer: Option D
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