G PREP 1
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- prachi18oct
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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!
- GMATGuruNY
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The portion in red undercounts the number of favorable outcomes.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!
Let the 4 letters be A, B, C and D.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
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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- GMATinsight
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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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