Gmat prep Question

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rakaisraka: Senior | Next Rank: 100 Posts; Posts: 37; Joined: Mon Jul 06, 2015 2:46 am

by rakaisraka » Wed Aug 26, 2015 11:53 am

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelop with its correct addresses. If the 4 leters are to be put into the 4 envelops at random , what is the probability that only 1 letter will be put into the envelop with its correct address.
a) 1/24
b) 1/8
c) 1/4
d) 1/3
e) 3/8

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Source: Beat The GMAT — Problem Solving | Wed Aug 26, 2015 11:53 am

kvamsy: Senior | Next Rank: 100 Posts; Posts: 52; Joined: Tue Mar 25, 2008 9:00 pm

by kvamsy » Wed Aug 26, 2015 12:32 pm

I think, Total ways of selecting the envolpes and letters are 4*4 ways.

Not getting the right letter into the envolope is 4+3+2+1 ways into fowur letters.

Total probability of not getting the letter into rt envolope is 10/16 ways.

Probability of getting rt letter into the rt envolope is 1 - 10/16 = 6/16, 3/8

Correct me if i am wrong.

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Max@Math Revolution: Elite Legendary Member; Posts: 3991; Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members; Location: Las Vegas, USA; Thanked: 19 times

by Max@Math Revolution » Thu Aug 27, 2015 3:18 am

Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelop with its correct addresses. If the 4 leters are to be put into the 4 envelops at random , what is the probability that only 1 letter will be put into the envelop with its correct address.
a) 1/24
b) 1/8
c) 1/4
d) 1/3
e) 3/8

==> probability = number of specific event / total number of cases

Total number of cases = letters(A,B,C,D), envelops(a,b,c,d) thus becomes 4*3*2*1.(A has 1 to choose from a,b,c,d, making it 4, while B has 3, C has 2 and D have 1)

The specific event has the cases (A,a) with (B,d) for (A,a), (C,b), (D,c) or (B,c), (C,d), (D,b), thus there is only 2 case (Since there should be only one letter in the envelope with the exact address)

Since the case applies to (B,b), (C,c), (D,d) as well, the number of cases that only 1 letter will be in the envelope with the exact address =2*4=8.

Thus, probability = 8/4*3*2*1=1/3. Therefore the answer is D.

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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 Aug 27, 2015 3:37 am

rakaisraka wrote:Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelop with its correct addresses. If the 4 leters are to be put into the 4 envelops 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

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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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 Dec 29, 2019 9:53 am

rakaisraka wrote:Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelop with its correct addresses. If the 4 leters are to be put into the 4 envelops at random , what is the probability that only 1 letter will be put into the envelop with its correct address.
a) 1/24
b) 1/8
c) 1/4
d) 1/3
e) 3/8

Let's solve this question using counting methods.

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.

-------------------
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
-------------------

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.
-------------------

So, P(exactly one letter with correct address) = 8/24 = 1/3

Answer: D

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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