GMAT PREP PS Problem

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alex.gellatly: Master | Next Rank: 500 Posts; Posts: 435; Joined: Wed Nov 16, 2011 7:27 amFollowed by: 16 members; Thanked: 48 times

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

Thanks

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Source: Beat The GMAT — Problem Solving | Sat Apr 21, 2012 3:56 am

Shalabh's Quants: Master | Next Rank: 500 Posts; Posts: 134; Joined: Fri Apr 06, 2012 3:11 amFollowed by: 5 members; Thanked: 35 times

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.

Shalabh Jain,
e-GMAT Instructor

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

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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nita24: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Mon Feb 06, 2012 8:44 am

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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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

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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LennonB: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Sun Oct 07, 2012 6:36 am

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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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

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)

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

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

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saurabhdhakad: Junior | Next Rank: 30 Posts; Posts: 29; Joined: Mon Jan 16, 2012 9:22 pm; Location: India; Thanked: 1 times

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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vinay1983: Legendary Member; Posts: 643; Joined: Wed Aug 14, 2013 4:27 amFollowed by: 7 members; Thanked: 48 times

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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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

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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