working with logic rather than math, (let capitals be the letters and lower case be the envelopes)
1. three possibilities for letter A = Aa, Ab, Ac
2. you then have 2 envelopes and 2 letters left. There are only two ways to arrange those:
e.g. Aa Bb Cc or Aa Bc Cb.
Two possibilities for BC given each of three possibilities for A = 3x2 = 6
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crackinggmat
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Tani Wolff - Kaplan wrote:The probability of getting the first one wrong is 2/3.
You now have two letters left and two envelopes. The probability of getting the second one wrong is 1/2.
Since you only have one envelope left, the probability on the third is 1.
Multiplying these three probabilities you get 2/3*1/2*1 = 1/3
Another way to look at it:
Three total possibilities, one favorable:
1. r,r,r
2. w,w,w
3. w,w,r
(Note: it is impossible to get r,r,w)
Hi Tani
can u pls help me pick my mistake here....
probability to put all letters in their correct envelope is 1/3 * 1/2 * 1/1 = 1/6
therefore probability of not getting any of these letters in the corrcet envelope is 1 - 1/6 = 5/6
Hi
In this cant we do 1-P(All reach the right envelope) ie 1-1/6=5/6?
When do we normally apply the 1- P rule apart from atleast cases?
In this cant we do 1-P(All reach the right envelope) ie 1-1/6=5/6?
When do we normally apply the 1- P rule apart from atleast cases?
Tani wrote:The probability of getting the first one wrong is 2/3.
You now have two letters left and two envelopes. The probability of getting the second one wrong is 1/2.
Since you only have one envelope left, the probability on the third is 1.
Multiplying these three probabilities you get 2/3*1/2*1 = 1/3
Another way to look at it:
Three total possibilities, one favorable:
1. r,r,r
2. w,w,w
3. w,w,r
(Note: it is impossible to get r,r,w)
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Hi sinhap07,
Yes - you can use the 1 - P(of what you're NOT looking for). However, you have not considered all of the options that fit that description.
Since the question asks for the probability that NONE of the addressees receive the proper letter, you have to account for:
1) the probability of ALL the letters getting to the proper addressee
2) the probability of just 1 of the letters getting to the proper addressee
Using this math approach, the calculation would be 1 - (1/6 + 1/2) = 1 - (4/6) = 2/6 = 1/3
GMAT assassins aren't born, they're made,
Rich
Yes - you can use the 1 - P(of what you're NOT looking for). However, you have not considered all of the options that fit that description.
Since the question asks for the probability that NONE of the addressees receive the proper letter, you have to account for:
1) the probability of ALL the letters getting to the proper addressee
2) the probability of just 1 of the letters getting to the proper addressee
Using this math approach, the calculation would be 1 - (1/6 + 1/2) = 1 - (4/6) = 2/6 = 1/3
GMAT assassins aren't born, they're made,
Rich
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This question was modeled on the following official question: https://www.beatthegmat.com/gmat-prep-ps ... 10570.html
Worth checking out for additional practice
Worth checking out for additional practice
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Scott@TargetTestPrep
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We are given that there are 3 letters and 3 envelopes. We must determine the probability that no letter is put in the correct envelope. Let's start with letter one.ruplun wrote:If 3 letters are put in 3 envelopes with three different adresses , what is the probability that no addressee receives the correct letter?
a. 1/6
b. 1/4
c. 1/3
d. 1/2
e. 2/3
Since there are 3 letters and only one correct envelope for letter one (which means two incorrect envelopes), the probability that letter one is put in the incorrect envelope is 2/3. Since there are 2 letters left and only one correct envelope for letter two (which means one incorrect envelope left), the probability that letter two is put in the incorrect envelope is 1/2. This also means that there is a 1/1 chance that the third letter is put in an incorrect envelope.
Thus, the probability that no letters are put in their correct envelopes is 2/3 x 1/2 x 1/1 = 1/3.
Answer: C
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Worst case, you can always write out each scenario.
Suppose we've got letters A, B, and C, and they're going to people a, b, and c.
If people a, b, and c receive the letters in that order (left to right), we've got the following ways of mailing:
ABC
ACB
BAC
BCA
CAB
CBA
The first scenario gives everyone the correct letter, the second gives A the correct letter, the third gives C the correct letter, the fourth (BAC) gives NO ONE the correct letter, the fifth (CAB) gives NO ONE the correct letter, and the sixth gives B the correct letter.
So 2 of our 6 fail to deliver anyone the correct letter, and 2/6 = 1/3.
Suppose we've got letters A, B, and C, and they're going to people a, b, and c.
If people a, b, and c receive the letters in that order (left to right), we've got the following ways of mailing:
ABC
ACB
BAC
BCA
CAB
CBA
The first scenario gives everyone the correct letter, the second gives A the correct letter, the third gives C the correct letter, the fourth (BAC) gives NO ONE the correct letter, the fifth (CAB) gives NO ONE the correct letter, and the sixth gives B the correct letter.
So 2 of our 6 fail to deliver anyone the correct letter, and 2/6 = 1/3.
