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Ida had 5 cards with matching envelopes - same in design, different in

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Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   22 May 2018, 22:44
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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
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C
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Bunuel
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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   23 May 2018, 00:05
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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


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https://gmatclub.com/forum/letter-arrang ... 84912.html (All possible cases)
https://gmatclub.com/forum/micky-has-10- ... 13801.html
https://gmatclub.com/forum/5-letters-hav ... 89501.html
https://gmatclub.com/forum/5-letters-ha ... 14675.html

Hope it helps.
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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   23 May 2018, 00:01
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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


Answer: D

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 \(= (1/5)*(3/4)*(2/3)*(1/2)*(1) * (5!/4!) = 1/4\)
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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   14 Apr 2019, 14:31
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Bunuel , VeritasKarishma, mikemcgarry

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
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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   15 Apr 2019, 01:44
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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
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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   15 Apr 2019, 04:57
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sayan640 wrote:
Bunuel , VeritasKarishma, mikemcgarry

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.
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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   Updated on: 15 Apr 2019, 05:39
VeritasKarishma wrote:
sayan640 wrote:
Bunuel , VeritasKarishma, mikemcgarry

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.
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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   15 Apr 2019, 05:35
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sayan640 wrote:
VeritasKarishma wrote:
sayan640 wrote:
Bunuel , VeritasKarishma, mikemcgarry

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.
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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   15 Apr 2019, 05:54
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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

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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   15 Apr 2019, 07:32
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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]


VeritasKarishma , sayan640
hi could you please advise on the highlighted part. How did we arrive at 9?
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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   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!)

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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   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.
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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   15 Apr 2019, 07:44
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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!)

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akdebonair well I have never come across this formula in probability of de arrangement , anyways thanks for the explanation
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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   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]




VeritasKarishma , sayan640
hi could you please advise on the highlighted part. How did we arrive at 9?[/quote]

Hi Archit Archit3110,

Please go through this...

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/blog/2011/ ... envelopes/
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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   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.
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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   15 Apr 2019, 10:56
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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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
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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   15 Apr 2019, 11:01
Yes understood where I went wrong.. thank you sayan640

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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   15 Apr 2019, 20:09
Other than derangement do we have any other method of solving this?

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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   15 Apr 2019, 22:51
sumi747
Did you find it difficult to understand anywhere ?

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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink] New post   15 Apr 2019, 22:53
sumi747 wrote:
Other than derangement do we have any other method of solving this?

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sumi747 please see my solution , I have used method given in Veritas link.
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Re: Ida had 5 cards with matching envelopes - same in design, different in [#permalink]
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