Hey,
I have no clue how to solve this Q.
Probability
This topic has expert replies
- [email protected]
- Master | Next Rank: 500 Posts
- Posts: 218
- Joined: Wed Dec 11, 2013 4:02 am
- Thanked: 3 times
- Followed by:4 members
- theCodeToGMAT
- Legendary Member
- Posts: 1556
- Joined: Tue Aug 14, 2012 11:18 pm
- Thanked: 448 times
- Followed by:34 members
- GMAT Score:650
A B C D
To find: only 1 letter into correct
probability of A's correct = 1/4 * 2/3 * 1/2 * 1 = 1/12
This is same for all
So,
4 * 1/12 = 1/3
{D}
To find: only 1 letter into correct
probability of A's correct = 1/4 * 2/3 * 1/2 * 1 = 1/12
This is same for all
So,
4 * 1/12 = 1/3
{D}
R A H U L
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Approach 1: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
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.
The correct answer is D.
Approach 2:
Let the correct order for the letters be A-B-C-D.
Case 1: Only A in the correct envelope
P(A is placed in the CORRECT envelope) = 1/4. (Of the 4 envelopes, 1 is correct.)
P(B is placed in an INCORRECT envelope) = 2/3. (Of the 3 remaining envelopes, only 1 is correct, so 2 are incorrect.)
At this point, B has taken either the envelope for C or the envelope for D.
Thus, of C and D, ONLY ONE could now be placed in the correct envelope.
P(this letter is placed in an INCORRECT envelope) = 1/2. (Of the 2 remaining envelopes, only 1 is correct, so 1 is incorrect.)
P(last letter is placed in an INCORRECT envelope) = 1. (As noted above, only ONE of C and D could be placed in the correct envelope, so the other must be placed in an incorrect envelope.)
Since all of these events must happen in order for A to be the only envelope correctly placed, we multiply the fractions:
1/4 * 2/3 * 1/2 * 1.
Other cases:
Since there are 4 options for the one envelope correctly placed -- A, B, C, or D -- the result above must be multiplied by 4:
1/4 * 2/3 * 1/2 * 1 * 4 = 1/3.
Approach 3:
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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
- [email protected]
- Master | Next Rank: 500 Posts
- Posts: 218
- Joined: Wed Dec 11, 2013 4:02 am
- Thanked: 3 times
- Followed by:4 members
HEy Guru!GMATGuruNY wrote:Approach 1: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
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.
The correct answer is D.
Approach 2:
Let the correct order for the letters be A-B-C-D.
Case 1: Only A in the correct envelope
P(A is placed in the CORRECT envelope) = 1/4. (Of the 4 envelopes, 1 is correct.)
P(B is placed in an INCORRECT envelope) = 2/3. (Of the 3 remaining envelopes, only 1 is correct, so 2 are incorrect.)
At this point, B has taken either the envelope for C or the envelope for D.
Thus, of C and D, ONLY ONE could now be placed in the correct envelope.
P(this letter is placed in an INCORRECT envelope) = 1/2. (Of the 2 remaining envelopes, only 1 is correct, so 1 is incorrect.)
P(last letter is placed in an INCORRECT envelope) = 1. (As noted above, only ONE of C and D could be placed in the correct envelope, so the other must be placed in an incorrect envelope.)
Since all of these events must happen in order for A to be the only envelope correctly placed, we multiply the fractions:
1/4 * 2/3 * 1/2 * 1.
Other cases:
Since there are 4 options for the one envelope correctly placed -- A, B, C, or D -- the result above must be multiplied by 4:
1/4 * 2/3 * 1/2 * 1 * 4 = 1/3.
Approach 3:
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.
Perfect explanation.
Thanx
Regards,
Mukherjee