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06 Oct 2009, 14:41
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There was a topic with problem:
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 in 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?
And seems that it was confusing for many. GMAT often has similar questions, so find below the problems to master yourself in them.
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 in 4 envelopes at random, what is the probability:
A. That no letter will be put into the envelope with its correct address?
B. That all letters will be put into the envelope with its correct address?
C. That only 1 letter will be put into the envelope with its correct address?
D. That only 2 letters will be put into the envelope with its correct address?
E. That only 3 letters will be put into the envelope with its correct address?
F. That more than one letter will be put into the envelope with its correct address?
G. That more than two letters will be put into the envelope with its correct address?
Note that each Q could be solved in different ways, so check your answers with the alternate solution.
Answers to follow after discussion.
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 in 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?
And seems that it was confusing for many. GMAT often has similar questions, so find below the problems to master yourself in them.
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 in 4 envelopes at random, what is the probability:
A. That no letter will be put into the envelope with its correct address?
B. That all letters will be put into the envelope with its correct address?
C. That only 1 letter will be put into the envelope with its correct address?
D. That only 2 letters will be put into the envelope with its correct address?
E. That only 3 letters will be put into the envelope with its correct address?
F. That more than one letter will be put into the envelope with its correct address?
G. That more than two letters will be put into the envelope with its correct address?
Note that each Q could be solved in different ways, so check your answers with the alternate solution.
Answers to follow after discussion.
10 Mar 2010, 04:39
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arjunsridhar84
Counting 1 correct: 4C1=4, choosing which letter will be put in correct envelope. (Let's assume A letter was put in correct envelope).
Then 3 letters (BCD) and 3 envelopes (BCD) are left. # of ways to put them incorrectly is 2:
Envelopes: B-C-D
Letters: C-D-B
OR: D-B-C
So total # of ways one letter will be put into the envelope with its correct address is: 4C1*2=8. As total # of ways to distribute 4 letter in 4 envelopes is 4!=24, probability P(C=1)=8/24.
Counting 2 correct: 4C2=6, choosing which 2 letter will be put in correct envelope. (Let's assume A and B letter were put in correct envelopes).
Then 2 letters (CD) and 2 envelopes (CD) are left. # of ways to put them incorrectly is 1:
Envelopes: C-D
Letters: D-C
So total # of ways two letters will be put into the envelope with its correct address is: 4C2*1=6. As total # of ways to distribute 4 letter in 4 envelopes is 4!=24, probability P(C=2)=6/24.
Counting 3 correct: if three letters will be put in correct envelopes, then the fourth one also gets its correct envelope. So there is no way we can put exactly 3 letters in correct envelopes --> P(C=3)=0
Counting 4 correct: there are 4!=24 ways to distribute 4 letters in 4 envelopes and obviously only one is when all letters get their correct envelopes. P(C=4)=1/24.
Counting all incorrect, or 0 correct:
P(all incorrect)=1-(1 correct)-(2 correct)-(3 correct)-(4 correct)=1-8/24-6/24-0-1/24=9/24.
ichha148
Opposite of "all letters in correct envelopes" (= 1-1/24 = 23/24), is: "at least one in wrong envelope" = "3 correct" (=0) + "2 correct" (=6/24) + "1 correct" (=8/24) + "0 correct, option A" (=9/24) --> \(1-\frac{1}{24}=\frac{23}{24}=0+\frac{6}{24}+\frac{8}{24}+\frac{9}{24}=\frac{23}{24}\).
Hope it's clear.
08 Oct 2009, 17:30
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yangsta8
Very good. Though there are some incorrect answers:
A.B. C. D. and E. correct. As for E: probability that only 3 letters will be put into the envelope with its correct address, is 0 because if you put 3 into the correct address envelopes, 4th one also gets the correct envelope.
F. Probability that more than one letter will be put into the envelope with its correct address is P(C=2)=6/24 plus P(C=3)=0 plus P(C=4)=1/24, which gives us P(C>1)=7/24
(You forgot to deduct P(C=1) in the way you were doing it)
G. Probability that more than two letters will be put into the envelope with its correct address is P(C=3)=0 plus P(C=4)=1/24, which gives us P(C>2)=1/24, the same probability as for P(C=4), because P(C=3)=0.
+1 to you.
This is a great post. 
