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GMAT PREP PS Problem

by alex.gellatly » Sat Apr 21, 2012 3:56 am
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

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by Shalabh's Quants » Sat Apr 21, 2012 5:51 am
alex.gellatly wrote: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

Thanks
Say Ist letter goes in correct envelope and others go in wrong envelopes, then

=> Prob. for for I letter going in correct envelope = 1/4;

=> Prob. for for II letter going in wrong envelope = 2/3;

=> Prob. for for III letter going in wrong envelope = 1/2;

=> Prob. for for IV letter going in wrong envelope = 1; ( only 1 wrong addressed envelope is left);

This event can occur with other 3 envelopes too.

Hence total prob. = 4*(1/4*2/3*1/2*1) = 1/3.
Last edited by Shalabh's Quants on Sat Apr 21, 2012 6:34 am, edited 1 time in total.
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by Anurag@Gurome » Sat Apr 21, 2012 6:18 am
alex.gellatly wrote: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

Thanks
Let the envelopes be E1, E2, E3, E4 and the corresponding letters be l1, l2, l3, l4.
Suppose l1 goes to E1 and the other letters do not go in their corresponding envelopes.
So, E2 will have either l3 or l4
If E2 has l3, E3 will have l4, E4 will have l2.
If E2 has l4, E3 will have l2, E4 will have l3.
So, for l1 going to E1 we have 2 arrangements where other letters are not going into right envelopes.
Similarly for l2 going to E2 or l3 going to E3 or l4 going to E4, we have 2 arrangements each.
Or there are 2*2*2*2 = 8 ways in which only one letter goes to right envelope.
Total number of all possible arrangements of letters in any envelopes is 4! = 24.

Required probability is 8/24 = [spoiler]1/3[/spoiler]

The correct answer is D.
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by nita24 » Wed Sep 19, 2012 11:16 pm
I have a doubt. I may be wrong but your help is required.
According to me this question can be solved as above:
probability = ((probability of putting A into A) * (probability of not putting B into B) * (probability of not putting C into C) * (probability of not putting D into D)) +
((probability of not putting A into A) * (probability of putting B into B) * (probability of not putting C into C) * (probability of not putting D into D)) +
((probability of not putting A into A) * (probability of not putting B into B) * (probability of putting C into C) * (probability of not putting D into D)) +
((probability of putting A into A) * (probability of not putting B into B) * (probability of not putting C into C) * (probability of putting D into D))
When i calculate probability in above manner the answer doesn't match with the OA.
Please help.
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by GMATGuruNY » Thu Sep 20, 2012 2:02 am
alex.gellatly wrote: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

Thanks
A DERANGEMENT is a permutation in which NO element is in the correct position.
Given n elements (where n>1):

Number of derangements = n! (1/2! - 1/3! + 1/4! + ... + ((-1)^n)/n!)

Let the 4 letters be ABCD.

Total arrangements:
Total number of ways to arrange the 4 letters = 4! = 24.

Good arrangements:
In a good arrangement, EXACTLY ONE letter is in the correct position.
Number of options for the one letter put into the correct position = 4. (A, B, C, or D)
Number of ways to DERANGE the 3 remaining letters = 3! (1/2! - 1/3!) = 3-1 = 2.
To combine these options, we multiply:
4*2 = 8.

Good/total = 8/24 = 1/3.

The correct answer is D.
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by LennonB » Sun Oct 07, 2012 7:20 am
This is driving me nuts. When I write out all the possibilities, I see only 28 possibilities (only 9 for all wrong: 2143, 2341, 2413, 3142, 3412, 3421, 4123 4312, 4321; 12 for exactly one correct; 6 for exactly 2 correct; ZERO for exactly 3 correct, and 1 for all correct). Thus, the probability of exactly 1 correct letter should be: 12/28 or 3/7. PLEASE tell me where I went wrong!

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

This is driving me nuts. When I write out all the possibilities, I see only 28 possibilities (only 9 for all wrong: 2341, 241
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by Brent@GMATPrepNow » Sun Oct 07, 2012 8:10 am
LennonB wrote:This is driving me nuts. When I write out all the possibilities, I see only 28 possibilities (only 9 for all wrong: 2143, 2341, 2413, 3142, 3412, 3421, 4123 4312, 4321; 12 for exactly one correct; 6 for exactly 2 correct; ZERO for exactly 3 correct, and 1 for all correct). Thus, the probability of exactly 1 correct letter should be: 12/28 or 3/7. PLEASE tell me where I went wrong!
I believe the problem (above in blue)is with your calculation of the number of possibilities with exactly one correct. Try listing them. I think you'll find that there are only 8 possibilities (not 12 as you have suggested)

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by Anurag@Gurome » Sun Oct 07, 2012 8:46 pm
alex.gellatly wrote: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

Thanks
Let the envelopes be E1, E2, E3, E4 and the corresponding letters be l1, l2, l3, l4.
Suppose l1 goes to E1 and the other letters do not go in their corresponding envelopes.
So, E2 will have either l3 or l4
If E2 has l3, E3 will have l4, E4 will have l2.
If E2 has l4, E3 will have l2, E4 will have l3.
So, for l1 going to E1 we have 2 arrangements where other letters are not going into right envelopes.
Similarly for l2 going to E2 or l3 going to E3 or l4 going to E4, we have 2 arrangements each.
Or there are 2*2*2*2 = 8 ways in which only one letter goes to right envelope.
Total number of all possible arrangements of letters in any envelopes is 4! = 24.

Required probability is 8/24 = [spoiler]1/3 [/spoiler]

The correct answer is D.
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by saurabhdhakad » Wed Aug 21, 2013 9:28 am
Hi Anurag,

Thanks for the easy way to solve this confusing arrangement.
Just wanted to clarify :
By 2*2*2*2 you mean 2+2+2+2 right? 8 total possible events.
i.e. when E1-L1 or E2-L2 or E3-L3 or E4 -L4 , so all possible events are Or-ed.

2*2*2*2 = 8 is slightly confusing.
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by vinay1983 » Wed Aug 21, 2013 8:37 pm
I am still confused about the solutions. :(

Not able to comprehend this.
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by GMATGuruNY » Wed Aug 21, 2013 8:56 pm
vinay1983 wrote:I am still confused about the solutions. :(

Not able to comprehend this.
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.
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