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All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2

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Joined: 22 Sep 2010
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All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink]

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22 Oct 2010, 05:52
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Difficulty:

95% (hard)

Question Stats:

31% (01:56) correct 69% (02:21) wrong based on 83 sessions

A.1/4
B.2/7
C.3/7
D.1/2
E.2/3

There are total of 2+2+2+2=8 bills. The sum of 2 bills will be 20 or more if we pick two $10 bills or$20 bill plus any bill for the other one (which means $20 bill and not$20 bill and two $20 dollar bills). $$\frac{2}{8}*\frac{1}{7}+2*\frac{2}{8}*\frac{6}{7}+\frac{2}{8}*\frac{1}{7}=\frac{1}{2}$$ (we multiply by 2 the second term as it represents the probability of$20/Not $20 scenario which can occur in 2 ways$20/Not $20 and Not$20/$20). Answer: D. Or another way: Find the probability of opposite event and subtract it from 1. Opposite event would be the sum being less than$20 which means the probability of picking any 2 bills but twenties (so any from 6) minus the case when we pick two $10 bills (1 case): $$\frac{C^2_6-1}{C^2_8}=\frac{1}{2}$$. Answer: D. _________________ Kudos [?]: 129172 [0], given: 12194 Intern Joined: 03 Sep 2010 Posts: 17 Kudos [?]: 13 [1], given: 0 Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink] Show Tags 24 Oct 2010, 12:30 1 This post received KUDOS Probabale combination for the sum >= 20 are (1,20) (20,1) , (5,20) , (20,5) , (20,10) , (10,20) , (10,10) , (20,20) For the first six combination let us calculate the individual probability = (2/8*2/7)*6 = 24/56 For the last two combination , the probabilty is (2/8*1/7)*2 = 4/56 Since the events/combinations are all mutually exclusive events Required Probability = 28/56 = 1/2 Kudos [?]: 13 [1], given: 0 GMAT Club Legend Joined: 09 Sep 2013 Posts: 16587 Kudos [?]: 273 [0], given: 0 Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink] Show Tags 12 Sep 2017, 00:53 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Kudos [?]: 273 [0], given: 0 Intern Joined: 25 Aug 2017 Posts: 1 Kudos [?]: 0 [0], given: 0 Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink] Show Tags 17 Sep 2017, 08:09 1 This post was BOOKMARKED Hi, I'm a little confused with this one. I understand there are 8 possible selections that can be made. So selection of the non (20,20) and (10,10) options, we have- 6*2/8...why multiply by 1/7? And I didn't understand the calculation for selecting (20,20) and (10,10). Any help will be appreciated. Thanks. Kudos [?]: 0 [0], given: 0 Math Expert Joined: 02 Sep 2009 Posts: 41892 Kudos [?]: 129172 [0], given: 12194 Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink] Show Tags 17 Sep 2017, 08:54 ragini246 wrote: Hi, I'm a little confused with this one. I understand there are 8 possible selections that can be made. So selection of the non (20,20) and (10,10) options, we have- 6*2/8...why multiply by 1/7? And I didn't understand the calculation for selecting (20,20) and (10,10). Any help will be appreciated. Thanks. {1, 1, 5, 5, 10, 10, 20, 20} P(20, not-20) = 2/8*6/7*2. We multiply by 2, (20, not-20) can occur in two ways: (20, not-20) and (not-20, 20) P(20, 20) = 2/8*1/7. _________________ Kudos [?]: 129172 [0], given: 12194 Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 1648 Kudos [?]: 848 [0], given: 3 Location: United States (CA) Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink] Show Tags 21 Sep 2017, 14:57 pzazz12 wrote: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 twenties. Two bills are drawn at random without replacement. What is the probability that their sum is$20 or more?

A. 1/4
B. 2/7
C. 3/7
D. 1/2
E. 2/3

The number of ways to select 2 bills from 8 bills is 8C2 = 8!/[2!(8-2)!] = (8 x 7)/2! = 28. Now we have to determine the number of ways to select 2 bills that total $20 or more. If one of the two bills is one of the$20 bills, then the sum is guaranteed to be $20 or more. Thus, for each$20 bill selected, we can pair it with the 6 non-$20 bills. Since there are two$20 bills, we have 6 x 2 = 12 pairings. Furthermore, we have two more pairings: we can pair the two $20 bills with each other and we can also pair the two$10 bills with each other. Thus, we have 14 pairings of two bills that total \$20 or more. Therefore, the probability is 14/28 = 1/2.

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Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2   [#permalink] 21 Sep 2017, 14:57
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