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ida had 5 cards with matching envelopes
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22 May 2018, 22:44
Question Stats:
18% (01:44) correct 82% (02:13) wrong based on 66 sessions
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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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Re: ida had 5 cards with matching envelopes
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23 May 2018, 00:01
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
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23 May 2018, 00:05
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/tanyaprepare ... 88626.htmlhttp://gmatclub.com/forum/letterarrang ... 84912.html (All possible cases) http://gmatclub.com/forum/mickyhas10 ... 13801.htmlhttp://gmatclub.com/forum/5lettershav ... 89501.htmlhttps://gmatclub.com/forum/5lettersha ... 14675.htmlHope it helps.
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Re: ida had 5 cards with matching envelopes
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14 Apr 2019, 14:31
Bunuel , VeritasKarishma, mikemcgarryTotal 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
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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 Globaltotal 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
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15 Apr 2019, 04:57
sayan640 wrote: Bunuel , VeritasKarishma, mikemcgarryTotal 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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ida had 5 cards with matching envelopes
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Updated on: 15 Apr 2019, 05:39
VeritasKarishma wrote: sayan640 wrote: Bunuel , VeritasKarishma, mikemcgarryTotal 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. BunuelPlease 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
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15 Apr 2019, 05:35
sayan640 wrote: VeritasKarishma wrote: sayan640 wrote: Bunuel , VeritasKarishma, mikemcgarryTotal 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
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15 Apr 2019, 05:54
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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ida had 5 cards with matching envelopes
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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) = 9So 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] BunuelPlease 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 , sayan640hi 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
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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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ida had 5 cards with matching envelopes
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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
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15 Apr 2019, 07:44
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



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ida had 5 cards with matching envelopes
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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] BunuelPlease 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 , sayan640hi could you please advise on the highlighted part. How did we arrive at 9?[/quote] Hi Archit Archit3110, Please go through this... https://www.veritasprep.com//blog/2011/ ... envelopes/



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ida had 5 cards with matching envelopes
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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
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15 Apr 2019, 10:56
Archit3110 wrote: sayan640 wrote: Archit3110 wrote: Total no of ways = 5! = 120 sayan640I 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



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Re: ida had 5 cards with matching envelopes
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15 Apr 2019, 11:01
Yes understood where I went wrong.. thank you sayan640Posted from my mobile device



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Re: ida had 5 cards with matching envelopes
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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
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15 Apr 2019, 22:51
sumi747 Did you find it difficult to understand anywhere ? Posted from my mobile device



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Re: ida had 5 cards with matching envelopes
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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
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