GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Oct 2019, 04:54

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# ida had 5 cards with matching envelopes

Author Message
TAGS:

### Hide Tags

Manager
Joined: 24 Jun 2013
Posts: 144
Location: India
Schools: ISB '20, GMBA '20

### Show Tags

22 May 2018, 22:44
10
00:00

Difficulty:

85% (hard)

Question Stats:

18% (01:44) correct 82% (02:13) wrong based on 66 sessions

### HideShow timer Statistics

Ida had 5 cards with matching envelopes - same in design, different in color. She removed the cards from all envelopes and randomly put them back. What is the probability that exactly one card got into the matching envelope?

A $$\frac{3}{16}$$
B $$\frac{5}{8}$$
C $$\frac{3}{8}$$
D $$\frac{1}{4}$$
E $$\frac{1}{2}$$

Source : Experts Global

_________________
If this post helped you learn something pls give kudos
Director
Joined: 14 Dec 2017
Posts: 516
Location: India

### Show Tags

23 May 2018, 00:01
1
3
doomedcat wrote:
ida had 5 cards with matching envelopes - same in design, different in color. She removed the cards from all envelopes and randomly put them back. What is the probability that exactly one card got into the matching envelope?

A $$\frac{3}{16}$$
B $$\frac{5}{8}$$
C $$\frac{3}{8}$$
D $$\frac{1}{4}$$
E $$\frac{1}{2}$$

Source : Experts Global

The question asks for "exactly one card into the right envelope", hence the combination of right & wrong cards will look like

R W W W W

Assuming the first card is put in right envelope, since there is only one right card out of the 5 for the first envelope, the probability is 1/5

For the second envelope, there are 3 wrong cards out of the 4 cards left, hence the probability for the Second card in wrong envelope is 3/4

For the third envelope, there are 2 wrong cards out of the 3 cards left, hence the probability for the third card in wrong envelope is 2/3

For the fourth envelope, there is 1 wrong card out of the 2 cards left, hence the probability for the fourth card in wrong envelope is 1/2

For the fifth envelope, since only one card is left now & a wrong one at that, hence the probability for the fifth card in wrong envelope is 1

Moreover the arrangement R W W W W can be arranged in 5!/4! ways

Hence the probability of exactly one right card $$= (1/5)*(3/4)*(2/3)*(1/2)*(1) * (5!/4!) = 1/4$$
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 58428

### Show Tags

23 May 2018, 00:05
1
2
doomedcat wrote:
ida had 5 cards with matching envelopes - same in design, different in color. She removed the cards from all envelopes and randomly put them back. What is the probability that exactly one card got into the matching envelope?

A $$\frac{3}{16}$$
B $$\frac{5}{8}$$
C $$\frac{3}{8}$$
D $$\frac{1}{4}$$
E $$\frac{1}{2}$$

Source : Experts Global

Similar questions to practice:
http://gmatclub.com/forum/tanya-prepare ... 88626.html
http://gmatclub.com/forum/letter-arrang ... 84912.html (All possible cases)
http://gmatclub.com/forum/micky-has-10- ... 13801.html
http://gmatclub.com/forum/5-letters-hav ... 89501.html
https://gmatclub.com/forum/5-letters-ha ... 14675.html

Hope it helps.
_________________
Manager
Joined: 29 Oct 2015
Posts: 226

### Show Tags

14 Apr 2019, 14:31
1

Total no of ways = 5! = 120

No of ways one card can be chosen which goes into the correct envelope out of 5 cards is = 5 c 1 = 5

The no of ways in which other four cards can be put into wrong envelopes = D (4) = 9
So the required probability = 5 * 9 / 120 = 45 / 120 = 3 /8
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5031
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

### Show Tags

15 Apr 2019, 01:44
doomedcat wrote:
Ida had 5 cards with matching envelopes - same in design, different in color. She removed the cards from all envelopes and randomly put them back. What is the probability that exactly one card got into the matching envelope?

A $$\frac{3}{16}$$
B $$\frac{5}{8}$$
C $$\frac{3}{8}$$
D $$\frac{1}{4}$$
E $$\frac{1}{2}$$

Source : Experts Global

total cards = 5
right ; 1/5
wrong ; 3/4*2/3*1/2 *1/1
total
1/5*3/4*2/3*1/2 *1/1 * 5c4 ways = 1/4
IMO D
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9701
Location: Pune, India

### Show Tags

15 Apr 2019, 04:57
sayan640 wrote:

Total no of ways = 5! = 120

No of ways one card can be chosen which goes into the correct envelope out of 5 cards is = 5 c 1 = 5

The no of ways in which other four cards can be put into wrong envelopes = D (4) = 9
So the required probability = 5 * 9 / 120 = 45 / 120 = 3 /8

This is correct.
Not sure what the official explanation says and why the official answer is (1/4) if it is indeed 1/4.
_________________
Karishma
Veritas Prep GMAT Instructor

Manager
Joined: 29 Oct 2015
Posts: 226

### Show Tags

Updated on: 15 Apr 2019, 05:39
sayan640 wrote:

Total no of ways = 5! = 120

No of ways one card can be chosen which goes into the correct envelope out of 5 cards is = 5 c 1 = 5

The no of ways in which other four cards can be put into wrong envelopes = D (4) = 9
So the required probability = 5 * 9 / 120 = 45 / 120 = 3 /8

This is correct.
Not sure what the official explanation says and why the official answer is (1/4) if it is indeed 1/4.

Bunuel
Please correct the OA...Unnecessarily giving way to confusion....Please change the OA from D to option C...You may refer to my explanation for OE...

Originally posted by sayan640 on 15 Apr 2019, 05:31.
Last edited by sayan640 on 15 Apr 2019, 05:39, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 58428

### Show Tags

15 Apr 2019, 05:35
sayan640 wrote:
sayan640 wrote:

Total no of ways = 5! = 120

No of ways one card can be chosen which goes into the correct envelope out of 5 cards is = 5 c 1 = 5

The no of ways in which other four cards can be put into wrong envelopes = D (4) = 9
So the required probability = 5 * 9 / 120 = 45 / 120 = 3 /8

This is correct.
Not sure what the official explanation says and why the official answer is (1/4) if it is indeed 1/4.

Bunuel
Please correct the OA...Unnecessarily giving way to confusion....Please change the OA from D to option C...You may refer to my explanation for OE...

________________________________
Done. Thank you.
_________________
Intern
Joined: 09 Mar 2019
Posts: 11

### Show Tags

15 Apr 2019, 05:54
1
Selecting one right envelope = 5 C 1 = 5 ways
Derangment of 4 envelopes = 4!(1- 1/1!+1/2!-1/3!+1/4!) = 9
Total ways of arrangement = 120

Probability of 1 envelope placed in right place = 9*5/120 = 3/8

Posted from my mobile device
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5031
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

### Show Tags

15 Apr 2019, 07:32
Total no of ways = 5! = 120

No of ways one card can be chosen which goes into the correct envelope out of 5 cards is = 5 c 1 = 5

The no of ways in which other four cards can be put into wrong envelopes = D (4) = 9
So the required probability = 5 * 9 / 120 = 45 / 120 = 3 /8[/quote]

This is correct.
Not sure what the official explanation says and why the official answer is (1/4) if it is indeed 1/4.[/quote]

Bunuel
Please correct the OA...Unnecessarily giving way to confusion....Please change the OA from D to option C...You may refer to my explanation for OE... [/quote]

hi could you please advise on the highlighted part. How did we arrive at 9?
Intern
Joined: 09 Mar 2019
Posts: 11

### Show Tags

15 Apr 2019, 07:40
It is formula for derangment derangment of n items is given by following formula.
D(n) = n!(1- 1/1!+1/2!-1/3!+1/4!......1/n!)

Posted from my mobile device
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5031
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

### Show Tags

Updated on: 16 Apr 2019, 00:25
Total no of ways = 5! = 120

my take on question below as per my understanding of the link
total cards given 5

so chances of selecting 1 correct card into 1 correct envelope = 5c1
total ways of choosing a card for an envelope = 5!
now we have to select ways to select a card which is to put into wrong envelope
so for 4 cards we have 3 ways to choose wrong envelope
for 3 cards we have 3 ways to choose wrong envelope
and for 2 cards we have 1 way to choose wrong envelope
total ways we can put 4 cards in wrong envelope; 3*3*1 = 9
The probability that exactly one card got into the matching envelope
5c1*9/5! = 3/8

Originally posted by Archit3110 on 15 Apr 2019, 07:42.
Last edited by Archit3110 on 16 Apr 2019, 00:25, edited 1 time in total.
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5031
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

### Show Tags

15 Apr 2019, 07:44
1
1
akdebonair wrote:
It is formula for derangment derangment of n items is given by following formula.
D(n) = n!(1- 1/1!+1/2!-1/3!+1/4!......1/n!)

Posted from my mobile device

akdebonair well I have never come across this formula in probability of de arrangement , anyways thanks for the explanation
Manager
Joined: 29 Oct 2015
Posts: 226

### Show Tags

15 Apr 2019, 09:16
Archit3110 wrote:
Total no of ways = 5! = 120

No of ways one card can be chosen which goes into the correct envelope out of 5 cards is = 5 c 1 = 5

The no of ways in which other four cards can be put into wrong envelopes = D (4) = 9
So the required probability = 5 * 9 / 120 = 45 / 120 = 3 /8

This is correct.
Not sure what the official explanation says and why the official answer is (1/4) if it is indeed 1/4.[/quote]

Bunuel
Please correct the OA...Unnecessarily giving way to confusion....Please change the OA from D to option C...You may refer to my explanation for OE... [/quote]

hi could you please advise on the highlighted part. How did we arrive at 9?[/quote]

Hi Archit Archit3110,

https://www.veritasprep.com//blog/2011/ ... envelopes/
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5031
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

### Show Tags

Updated on: 15 Apr 2019, 22:54
Total no of ways = 5! = 120

my take on question below as per my understanding of the link
total cards given 5

so chances of selecting 1 correct card into 1 correct envelope = 5c1
total ways of choosing a card for an envelope = 5!
now we have to select ways to select a card which is to put into wrong envelope
so for 4 cards we have 3 ways to choose wrong envelope
for 3 cards we have 3 ways to choose wrong envelope
and for 2 cards we have 1 way to choose wrong envelope
total ways we can put 4 cards in wrong envelope; 3*3*1 = 9
The probability that exactly one card got into the matching envelope
5c1*9/5! = 3/8

Originally posted by Archit3110 on 15 Apr 2019, 10:31.
Last edited by Archit3110 on 15 Apr 2019, 22:54, edited 2 times in total.
Manager
Joined: 29 Oct 2015
Posts: 226

### Show Tags

15 Apr 2019, 10:56
1
Archit3110 wrote:
sayan640 wrote:
Archit3110 wrote:
Total no of ways = 5! = 120

sayan640
I went through the link but I am not able to understand how you getting 9
my take on question below as per my understanding of the link
total cards given 5

so chances of selecting 1 correct card into 1 correct envelope = 5c1
total ways of choosing a card for an envelope = 5!
now we have to select ways to select a card which is to put into wrong envelope
so for 4 cards we have 3 ways to choose wrong envelope
for 3 cards we have 2 ways to choose wrong envelope
and for 2 cards we have 1 way to choose wrong envelope
total ways we can put 4 cards in wrong envelope; 3*2*1 = 6
The probability that exactly one card got into the matching envelope
5c1*6/5! = 1/4

VeritasKarishma sayan640 ; please explain /highlight where exactly am i going wrong..

Total no of ways = 5! = 120

No of ways one card can be chosen which goes into the correct envelope out of 5 cards is = 5 c 1 = 5

The no of ways in which other four cards can be put into wrong envelopes = D (4) = 9
( It has been derived in this link :- https://www.veritasprep.com//blog/2011/ ... envelopes/ )
( In question 2 ..no 5 ..to make your life easier..)
Now I think you understood how D(4) = 9..

So the required probability = 5 * 9 / 120 = 45 / 120 = 3 /8
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5031
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

### Show Tags

15 Apr 2019, 11:01
Yes understood where I went wrong.. thank you sayan640

Posted from my mobile device
Manager
Joined: 17 May 2018
Posts: 140
Location: India

### Show Tags

15 Apr 2019, 20:09
Other than derangement do we have any other method of solving this?

Posted from my mobile device
Manager
Joined: 29 Oct 2015
Posts: 226

### Show Tags

15 Apr 2019, 22:51
sumi747
Did you find it difficult to understand anywhere ?

Posted from my mobile device
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5031
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)

### Show Tags

15 Apr 2019, 22:53
sumi747 wrote:
Other than derangement do we have any other method of solving this?

Posted from my mobile device

sumi747 please see my solution , I have used method given in Veritas link.
Re: ida had 5 cards with matching envelopes   [#permalink] 15 Apr 2019, 22:53

Go to page    1   2    Next  [ 23 posts ]

Display posts from previous: Sort by