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. \(\dfrac1{24}\)
B. \(\dfrac18\)
C. \(\dfrac14\)
D. \(\dfrac13\)
E. \(\dfrac38\)
Answer: D
Source: GMAT Prep
Tanya prepared \(4\) different letters to be sent to \(4\) different addresses. For each letter, she prepared an envelop
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There are 4! or 24 total ways the letters can be distributed among the envelopes.
Put A in its correct envelope. B,C and D therefore must be put into incorrect envelopes.
B letter can be put into C envelope, leaving C and D letters and B and D envelopes. D must therefore go into B envelope and C into D envelope. So 1 way in this case.
By the same approach B can be put also into D envelope, leaving C and D letters and B and C envelopes. D must go into C and C into B. 1 more way.
Since there are 4 letters, by the logic above, there are 4x2 or 8 total ways to put one letter correctly and the others incorrectly.
Probability is therefore 8/24=
[spoiler]
D, 1/3
[/spoiler]
Put A in its correct envelope. B,C and D therefore must be put into incorrect envelopes.
B letter can be put into C envelope, leaving C and D letters and B and D envelopes. D must therefore go into B envelope and C into D envelope. So 1 way in this case.
By the same approach B can be put also into D envelope, leaving C and D letters and B and C envelopes. D must go into C and C into B. 1 more way.
Since there are 4 letters, by the logic above, there are 4x2 or 8 total ways to put one letter correctly and the others incorrectly.
Probability is therefore 8/24=
[spoiler]
D, 1/3
[/spoiler]
Last edited by regor60 on Thu Sep 30, 2021 8:25 am, edited 1 time in total.
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Let's solve this question using counting methods.Gmat_mission wrote: ↑Wed Sep 29, 2021 8:39 amTanya 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. \(\dfrac1{24}\)
B. \(\dfrac18\)
C. \(\dfrac14\)
D. \(\dfrac13\)
E. \(\dfrac38\)
Answer: D
Source: GMAT Prep
So, P(exactly one letter with correct address) = (number of outcomes in which one letter has correct address)/(total number of outcomes)
Let a, b, c and d represent the letters, and let A, B, C and D represent the corresponding addresses.
So, let's list the letters in terms of the order in which they are delivered to addresses A, B, C, and D.
So, for example, the outcome abcd would represent all letters going to their intended addresses.
Likewise, cabd represent letter d going to its intended address, but the other letters not going to their intended addresses.
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total number of outcomes
The TOTAL number of outcomes = the number of different ways we can arrange a, b, c, and d
RULE: We can arrange n different objects in n! ways.
So, we can arrange the four letters in 4! ways = 24 ways
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number of outcomes in which one letter has correct address
Now let's list all possible outcomes in which exactly ONE letter goes to its intended addresses:
- acbd
- adbc
- cbda
- dbac
Aside: At this point you might recognize that there are two outcomes for each arrangement in which one letter goes to its intended address
- dacb
- bdca
- bcad
- cabd
There are 8 such outcomes.
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So, P(exactly one letter with correct address) = 8/24 = 1/3
Answer: D
Cheers,
Brent