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There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back and then another card is drawn and put back. What is the probability that one of the cards was 5 if the sum of the two was 8? (A) 1 (B) \(\frac{2}{3}\) (C) \(\frac{1}{2}\) (D) \(\frac{2}{5}\) (E) \(\frac{1}{3}\) Source: GMAT Club Tests  hardest GMAT questions Answer:The probability is \(\frac{2}{5}\) . You have the possibilities of 2,6 3,5 4,45,3 6,2 Two out of these five combinations have 5's in them, so the probability that you have a 5 is 2/5. I counted the event {4,4} twice in the calculation, which makes a probability of 2/6=1/3 so E. Even if they are the same cards, I don't understand why we can't make a permutation of the cards #4, i.e. counting it twice. Can you please tell me where I am wrong?Thanks in advance .



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11 Oct 2008, 09:55
4,4 in the reverse order also does not make any diff of permutation. So it comes once only.
But this is gud question !



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12 Oct 2008, 08:18
its simple
total possible case : 5 and out of 5 only 2 case are tehre which has card'5' init
so 2/5...
Thanks



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23 May 2009, 03:26
sorry isn't there a mistake here... the options are 2,6 3,5 5,3 6,2 no 4's allowed because there is only one 4 card. correspondingly... if total is 8 there is a 2/4 chance there is a 5 in the group (either (3,5) or (5,3)) therefore there is 50% chance 1/2. Am i missing something?
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24 May 2009, 23:50
nightwing79 wrote: sorry isn't there a mistake here...
the options
are
2,6 3,5 5,3 6,2
no 4's allowed because there is only one 4 card.
correspondingly... if total is 8 there is a 2/4 chance there is a 5 in the group (either (3,5) or (5,3)) therefore there is 50% chance 1/2.
Am i missing something? Cards are returned after draw, therefore 4,4 is possible.
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25 Mar 2010, 04:17
echizen wrote: There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back and then another card is drawn and put back. What is the probability that one of the cards was 5 if the sum of the two was 8? (A) 1 (B) \(\frac{2}{3}\) (C) \(\frac{1}{2}\) (D) \(\frac{2}{5}\) (E) \(\frac{1}{3}\) Source: GMAT Club Tests  hardest GMAT questions Answer:Total ways to get sum 8 = 2,6 6,2 4,4 3,5 5,3 p(favorable) = 2/5 hence D



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Re: m05 #10 [#permalink]
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29 Mar 2011, 04:36
Total = 2 *(2,6) + 2 * (3,5) + (4,4) Favorable = (3,5) or (5,3) P = 2/5 Answer  D
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29 Mar 2011, 21:32
Can somebody will try to explain this problem with union and intersection method. Not through permutation and combination. probability of getting 5 in first draw is 1/6, then in second draw outcome must be 3. How we will calculate this must be event. The probability of getting 3 in the second draw is also 1/6, but I think we need to calculate here must be event dependent on the first event.



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29 Mar 2011, 22:51
Isn't the phrasing of this question misleading? Possibly it's just me because it seems that everyone else got it, but would someone pls explain to me how I'm reading this question wrong?
Since it asks "What is the probability that one of the cards was 5 if the sum of the two was 8?", the condition "if the sum of the two was 8" implies only 2 situations that this can occur. Then if this condition is true, "What is the probability that one of the cards was 5?". In both situations (i.e. 3+5, 5+3), it can only happy 2/4 times or 1/2, so shouldn't the answer be C?
If the question asked "what is the probability of drawing a combination which sums to 8 if one card is 5?", then I would agree, 2/5 is correct.
Where am I going wrong??



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29 Mar 2011, 23:31
woojet wrote: Isn't the phrasing of this question misleading? Possibly it's just me because it seems that everyone else got it, but would someone pls explain to me how I'm reading this question wrong?
Since it asks "What is the probability that one of the cards was 5 if the sum of the two was 8?", the condition "if the sum of the two was 8" implies only 2 situations that this can occur. Then if this condition is true, "What is the probability that one of the cards was 5?". In both situations (i.e. 3+5, 5+3), it can only happy 2/4 times or 1/2, so shouldn't the answer be C?
If the question asked "what is the probability of drawing a combination which sums to 8 if one card is 5?", then I would agree, 2/5 is correct.
Where am I going wrong?? "What is the probability that one of the cards was 5 if the sum of the two was 8?", I think, sum of 8 can be achieved by all the five possible combination. First, you have to fulfill the if clause after that only you can think about then clause. i.e if sum was 8 then what is the prob of 5.



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29 Mar 2011, 23:58
rajeshaaidu wrote: woojet wrote: Isn't the phrasing of this question misleading? Possibly it's just me because it seems that everyone else got it, but would someone pls explain to me how I'm reading this question wrong?
Since it asks "What is the probability that one of the cards was 5 if the sum of the two was 8?", the condition "if the sum of the two was 8" implies only 2 situations that this can occur. Then if this condition is true, "What is the probability that one of the cards was 5?". In both situations (i.e. 3+5, 5+3), it can only happy 2/4 times or 1/2, so shouldn't the answer be C?
If the question asked "what is the probability of drawing a combination which sums to 8 if one card is 5?", then I would agree, 2/5 is correct.
Where am I going wrong?? "What is the probability that one of the cards was 5 if the sum of the two was 8?", I think, sum of 8 can be achieved by all the five possible combination. First, you have to fulfill the if clause after that only you can think about then clause. i.e if sum was 8 then what is the prob of 5.  Oh ok thanks  seems so obvious now once you put it that way. I was reading it the other way around which was leading me to think of 2 sets instead of 5 sets. Thanks for the clarification.



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Re: m05 #10 [#permalink]
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02 Apr 2012, 06:00
echizen wrote: There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back and then another card is drawn and put back. What is the probability that one of the cards was 5 if the sum of the two was 8? (A) 1 (B) \(\frac{2}{3}\) (C) \(\frac{1}{2}\) (D) \(\frac{2}{5}\) (E) \(\frac{1}{3}\) Source: GMAT Club Tests  hardest GMAT questions Answer:The probability is \(\frac{2}{5}\) . You have the possibilities of 2,6 3,5 4,45,3 6,2 Two out of these five combinations have 5's in them, so the probability that you have a 5 is 2/5. I counted the event {4,4} twice in the calculation, which makes a probability of 2/6=1/3 so E. Even if they are the same cards, I don't understand why we can't make a permutation of the cards #4, i.e. counting it twice. Can you please tell me where I am wrong?Thanks in advance . Probablity is like a dreaded subject...even the simplest problem(now) looks like a major hurdle..... thanks for the nice explanation....
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Re: m05 #10 [#permalink]
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02 Apr 2012, 08:57
Why are we considering (5,3) and (6,2) again after having considered (3,5) and (2,6)? When the card after first draw is put back, how does it matter whether 5 appears in the first draw or the second?
If possible, please explain with a simple example. Thanks.



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02 Apr 2012, 22:17
tssambi wrote: Why are we considering (5,3) and (6,2) again after having considered (3,5) and (2,6)? When the card after first draw is put back, how does it matter whether 5 appears in the first draw or the second?
If possible, please explain with a simple example. Thanks. here actually the arrangement is important. You can see it this way you first draw a 5 u put it back and then another pick gives 3 you now have 8 now in second go you first get 3 and then 5 =>8 the thing here is, in how many events/ actions you can have 8, whether you pick 5 first or 3 both matters because it counts for the number of events/ actions performed to get 8. hope this clarifies..!!
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Re: m05 #10 [#permalink]
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03 Apr 2012, 05:04
tssambi wrote: Why are we considering (5,3) and (6,2) again after having considered (3,5) and (2,6)? When the card after first draw is put back, how does it matter whether 5 appears in the first draw or the second?
If possible, please explain with a simple example. Thanks. There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back and then another card is drawn and put back. What is the probability that one of the cards was 5 if the sum of the two was 8?(A) 1 (B) \(\frac{2}{3}\) (C) \(\frac{1}{2}\) (D) \(\frac{2}{5}\) (E) \(\frac{1}{3}\) What combinations of two cards are possible to total 8? (first card, second card): (6,2) (2,6) (5,3) (3,5) (4,4) – only 5 possible scenarios for sum to be 8. One from this 5 has already happened. Notice here that the case of (6,2) is different from (2,6), and similarly the case of (5,3) is different from (3,5).From this five cases, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5: P=2/5. Answer: D. Check: theprobabilitythatavisitoratthemallbuysapackof85523.html for more. Hope it's clear.
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Re: m05 #10 [#permalink]
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09 Jul 2012, 22:08
I did it this way  Probability of the picking the first card  5/6  ( The cards must either by 2,4,5,6,3)
Probability of picking second card  1/6  Because if you select any of the five cards in the first card , you can pick only its complement in the second card  Eg: 2 must have 6
So  5/6 * 1/6 = 5/36
I am still confused where I made a mistake  Could some one help me out ?



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10 Jul 2012, 02:24
arvind410 wrote: I did it this way  Probability of the picking the first card  5/6  ( The cards must either by 2,4,5,6,3)
Probability of picking second card  1/6  Because if you select any of the five cards in the first card , you can pick only its complement in the second card  Eg: 2 must have 6
So  5/6 * 1/6 = 5/36
I am still confused where I made a mistake  Could some one help me out ? What do you mean when you say that "the probability of picking the first card is 5/6"? 5/6 is the probability of picking any of the five particular cards from 6. For example 5/6 is the probability of picking either 2, 3, 4, 5, or 6. But why do you consider this case? The same question for the second card. Anyway, we are told that two cards were selected and the sum of these two cards is 8 (this scenario has already happened). The question then asks about the probability that one of the cards was 5. There are five cases possible the sum to be 8: (6,2); (2,6); (5,3); (3,5); (4,4). One from this 5 has already happened. From this five cases, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5: P=2/5. Hope it's clear.
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03 Apr 2013, 08:31
echizen wrote: There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back and then another card is drawn and put back. What is the probability that one of the cards was 5 if the sum of the two was 8? (A) 1 (B) \(\frac{2}{3}\) (C) \(\frac{1}{2}\) (D) \(\frac{2}{5}\) (E) \(\frac{1}{3}\) Source: GMAT Club Tests  hardest GMAT questions Answer:The probability is \(\frac{2}{5}\) . You have the possibilities of 2,6 3,5 4,45,3 6,2 Two out of these five combinations have 5's in them, so the probability that you have a 5 is 2/5. I counted the event {4,4} twice in the calculation, which makes a probability of 2/6=1/3 so E. Even if they are the same cards, I don't understand why we can't make a permutation of the cards #4, i.e. counting it twice. Can you please tell me where I am wrong?Thanks in advance . I went the long way to solve this problem. The question is looking for a combination of 3 and 5. Step 1: In the first draw what are the chances of drawing a 3 or a 5? Simple. There are 6 options and the chances that any of the two will come up is 2/6 = 1/3. Step 2: Once again we have 6 options but one of 3 or 5 must have been chosen on the way to getting an 8. So we need to draw 5 if 3 had been chosen or draw 3 if 5 had been chosen. This time around we go the combinatorics way. 3 or 5 have become one option. Therefore the chances of choosing either of them becomes 6!/2!4! = 1*2*3*4*5*6/1*2*1*2*3*4 = 15. Flip this over and you get a 1/15 chance of choosing 3 or 5, if one had been chosen before. Step 3: This is an AND decision in which both Steps 1 and 2 must occur so we can have a 5 and a 3. That means we add 1/3 to 1/15. 1/3 + 1/15 = 2/5.
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03 Apr 2013, 18:08
I don't quite get why people count 5,3 and 3,5 as different combinations. The question says nothing about order. The way I reason this, you can have the following combinations
cards 6 and 2
5 and 3
4 and 4
So if one of the cards is 5, the other is 3, regardless of the order that they were pulled out of the stack. that's 1 out 3 cards, no?



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04 Apr 2013, 04:13
Dixon wrote: I don't quite get why people count 5,3 and 3,5 as different combinations. The question says nothing about order. The way I reason this, you can have the following combinations
cards 6 and 2
5 and 3
4 and 4
So if one of the cards is 5, the other is 3, regardless of the order that they were pulled out of the stack. that's 1 out 3 cards, no? We have two cards, thus the case (first card)=5 and (second card)=3 is different from (first card)=3 and (second card)=5. The following post might help: m0571518.html#p1069296m0571518.html#p1103156
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