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Re M1108
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16 Sep 2014, 00:44
Official Solution:There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are randomly selected from the lot, what is the probability that the positive difference between the numbers on these cards is 3 or less? A. \(\frac{8}{15}\) B. \(\frac{9}{15}\) C. \(\frac{10}{15}\) D. \(\frac{12}{15}\) E. \(\frac{14}{15}\) The only pair which has the positive difference more than 3 is {2,6}: \(62=4 \gt 3\). Probability of this pair to be drawn is \(\frac{1}{C^2_6}\). So, the probability that the positive difference between the numbers on the cards is 3 or less is: \(P=1\frac{1}{C^2_6}=\frac{14}{15}\). Answer: E
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Re: M1108
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30 Sep 2015, 23:27
Hi Bunuel,
Doesn't the use of the phrase "positive difference between the numbers on these cards" mean that we need to eliminate the possibility of picking two 5s as well. Difference between two 5s is 0, which is technically not a difference at all.



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31 Aug 2016, 05:22
I think this is a highquality question and I agree with explanation.



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Re: M1108
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17 Dec 2016, 05:33
Bunuel wrote: Official Solution:
There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are randomly selected from the lot, what is the probability that the positive difference between the numbers on these cards is 3 or less?
A. \(\frac{8}{15}\) B. \(\frac{9}{15}\) C. \(\frac{10}{15}\) D. \(\frac{12}{15}\) E. \(\frac{14}{15}\)
The only pair which has the positive difference more than 3 is {2,6}: \(62=4 \gt 3\). Probability of this pair to be drawn is \(\frac{1}{C^2_6}\). So, the probability that the positive difference between the numbers on the cards is 3 or less is: \(P=1\frac{1}{C^2_6}=\frac{14}{15}\).
Answer: E Hi Bunnel, 6C2 is number of ways in which cards can be chosen, not pairs. there are 2 ways in which this particular pair of number can be chosen (6,2 or 2,6) So shouldn't it be 1  2/15 = 13/15??
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Re: M1108
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17 Dec 2016, 11:20
sidoknowia wrote: Bunuel wrote: Official Solution:
There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are randomly selected from the lot, what is the probability that the positive difference between the numbers on these cards is 3 or less?
A. \(\frac{8}{15}\) B. \(\frac{9}{15}\) C. \(\frac{10}{15}\) D. \(\frac{12}{15}\) E. \(\frac{14}{15}\)
The only pair which has the positive difference more than 3 is {2,6}: \(62=4 \gt 3\). Probability of this pair to be drawn is \(\frac{1}{C^2_6}\). So, the probability that the positive difference between the numbers on the cards is 3 or less is: \(P=1\frac{1}{C^2_6}=\frac{14}{15}\).
Answer: E Hi Bunnel, 6C2 is number of ways in which cards can be chosen, not pairs. Choosing randomly 2 cards from the set means choosing pairsthere are 2 ways in which this particular pair of number can be chosen (6,2 or 2,6) (6,2) and (2,6) are considered the same since we dont cover the order in countingSo shouldn't it be 1  2/15 = 13/15?? This should be 11/15=14/15
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Re: M1108
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08 Jun 2017, 09:52
Bunuel wrote: Official Solution:
There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are randomly selected from the lot, what is the probability that the positive difference between the numbers on these cards is 3 or less?
A. \(\frac{8}{15}\) B. \(\frac{9}{15}\) C. \(\frac{10}{15}\) D. \(\frac{12}{15}\) E. \(\frac{14}{15}\)
The only pair which has the positive difference more than 3 is {2,6}: \(62=4 \gt 3\). Probability of this pair to be drawn is \(\frac{1}{C^2_6}\). So, the probability that the positive difference between the numbers on the cards is 3 or less is: \(P=1\frac{1}{C^2_6}=\frac{14}{15}\).
Answer: E Bunuel, would you please elaborate more on your answer? I initially approached this via 6!/3! to get rid of the duplicates and this gives 120, but I believe this is wrong because the problem implies that ordering does not matter because the problem states 'positive difference'. However I do not understand 6C2. What about the duplicate 5s? A simpler example would be a list of numbers 2,3,3 and I am trying to choose two numbers. 3C2 would give me (2,3), (3,3) and (2,3) for 3 combinations, but should it not only be 2?



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Re: M1108
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22 Jul 2017, 10:29
It's not just the digits, but also the cards. 6 'distinct' cards.



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06 Aug 2017, 18:50
Total possibilities: 15. There is only 1 combination when positive difference is more than 3(6,2) so the result is: 151=14/15



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Re: M1108
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20 Sep 2017, 05:44
Bunuel wrote: There are six cards bearing numbers 2, 4, 5, 5, 5, 6. If two cards are randomly selected from the lot, what is the probability that the positive difference between the numbers on these cards is 3 or less?
A. \(\frac{8}{15}\) B. \(\frac{9}{15}\) C. \(\frac{10}{15}\) D. \(\frac{12}{15}\) E. \(\frac{14}{15}\) I have a couple of doubts while solving this question . 1. if its not given the cards are distinct ,do we assume they are? 2. while eliminating the number pairs why do we not consider 6,2?



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Re: M1108
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27 Oct 2017, 03:51
How order of selection is important in M1014 (Set SS consists of all prime integers less than 10. If a number is selected from set SS at random and then another number, not necessarily different, is selected from set SS at random, what is the probability that the sum of these numbers is odd?), and not important in this question? There could be two selections (2,6) and (6,2)



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Re: M1108
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17 Jan 2018, 01:01
Hi Bunuel, I have same doubt as posted in this thread. Doesn't the use of the phrase "positive difference between the numbers on these cards" mean that we need to eliminate the possibility of picking two 5s as well. Difference between two 5s is 0, which is technically not a difference at all.
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17 Jan 2018, 03:00
Thanks Bunuel for clarification.
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30 Aug 2018, 00:37
Hi Bunnel, I have a small doubt here, what if we manually counted the favorable outcomes? I tried doing this way, The no. of ways that we select 2 from 6 is 6c2 (this goes to the denominator) and now the numerator, here except the pair(6,2) every other pair has to be counted((2,4),(2,5),(4,2),(4,5),(4,6),(5,2),(5,4),(5,6),(6,5),(6,4)).So no. of favorable outcomes is 10 and the answer would be 10/15. Help me out here.
Thanks Rajesh Bhupathi



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Re: M1108
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30 Aug 2018, 00:43
Bhupathirajesh1 wrote: Hi Bunnel, I have a small doubt here, what if we manually counted the favorable outcomes? I tried doing this way, The no. of ways that we select 2 from 6 is 6c2 (this goes to the denominator) and now the numerator, here except the pair(6,2) every other pair has to be counted((2,4),(2,5),(4,2),(4,5),(4,6),(5,2),(5,4),(5,6),(6,5),(6,4)).So no. of favorable outcomes is 10 and the answer would be 10/15. Help me out here.
Thanks Rajesh Bhupathi Try again, now taking into the account that we have THREE cards with number 5.
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