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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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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
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Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink]
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22 Oct 2010, 06:35
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 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): \(\frac{C^2_61}{C^2_8}=\frac{1}{2}\). Answer: D.
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Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink]
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24 Oct 2010, 12:30
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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



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Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink]
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17 Sep 2017, 08:09
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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.



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All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink]
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17 Sep 2017, 08:54
ragini246 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
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, not20) = 2/8*6/7*2. We multiply by 2, (20, not20) can occur in two ways: (20, not20) and (not20, 20) P(20, 20) = 2/8*1/7.
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Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink]
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21 Sep 2017, 14:57
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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!(82)!] = (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. Answer: D
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Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink]
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28 Oct 2017, 04:19
ScottTargetTestPrep wrote: 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!(82)!] = (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. Answer: D Pairing it with 20 and one twice does not make sense right? its the same. similarly to all other non $20 bills



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Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink]
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04 Dec 2017, 21:20
Given: {1,1},{5,5},{10,10},{20,20} Total number ways of selecting 2 bills : 4C2 ( any 2 out of {1,5,10,20}} + 4 (one of {1,1},{5,5},{10,10},{20,20}) = 10 ways Total number of selecting two bills whose sum > 20$ = {10,10} + number of combinations of 20$ and one of {1,5,10,20} = 1 + 4 = 5 Required probability = 5/10 = 1/2 Hi Bunuel, do you see any flaw in above method, while finding total ways of drawing 2 bills, i am eliminating duplicate combinations.



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All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink]
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05 Dec 2017, 06:07
Quote: 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
{1, 1, 5, 5, 10, 10, 20, 20}
P(20, not20) = 2/8*6/7*2. We multiply by 2, (20, not20) can occur in two ways: (20, not20) and (not20, 20)
P(20, 20) = 2/8*1/7. Where does 1/7 and 6/7 come from? Can any1 explain please {1, 1, 5, 5, 10, 10, 20, 20}  8 numbersP(20, not20) = 2/8*6/7*2. We multiply by 2, (20, not20) can occur in two ways: (20, not20) and (not20, 20) There are 8 numbers. The probability of picking 20 is 2/8 (because there are two 20's in eight numbers). After we pick one number there are 7 left out of which there is now only on 20 left. The probability of picking not20 is 6/7. P(20, 20) = 2/8*1/7. Again: 8 numbers. The probability of picking 20 is 2/8. After we pick one number there are 7 left out of which there is now only on 20 left. The probability of picking 20 again is 1/7. Hope it's clear.



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Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink]
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05 Dec 2017, 06:15
lichting wrote: Quote: 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
{1, 1, 5, 5, 10, 10, 20, 20}
P(20, not20) = 2/8*6/7*2. We multiply by 2, (20, not20) can occur in two ways: (20, not20) and (not20, 20)
P(20, 20) = 2/8*1/7. Where does 1/7 and 6/7 come from? Can any1 explain please {1, 1, 5, 5, 10, 10, 20, 20}  8 numbersP(20, not20) = 2/8*6/7*2. We multiply by 2, (20, not20) can occur in two ways: (20, not20) and (not20, 20) There are 8 numbers. The probability of picking 20 is 2/8 (because there are two 20's in eight numbers). After we pick one number there are 7 left out of which there is now only on 20 left. The probability of picking not20 is 6/7. P(20, 20) = 2/8*1/7. Again: 8 numbers. The probability of picking 20 is 2/8. After we pick one number there are 7 left out of which there is now only on 20 left. The probability of picking 20 again is 1/7. Hope it's clear. 22. Probability For more: ALL YOU NEED FOR QUANT ! ! !Ultimate GMAT Quantitative MegathreadHope it helps.
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Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink]
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05 Dec 2017, 06:39
Quote: P(20, 20) = 2/8*1/7. Again: 8 numbers. The probability of picking 20 is 2/8. After we pick one number there are 7 left out of which there is now only on 20 left. The probability of picking 20 again is 1/7. Thanks Bunnel. One more question, since the order doesn't matter here, do we need to multiple 2/8 * 1/7 * 2 like the P(20,not20)?



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Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink]
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