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

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Manager
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, 04:52
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Difficulty:

95% (hard)

Question Stats:

36% (01:32) correct 64% (02:05) wrong based on 142 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 [?]: 139148 [0], given: 12777 Intern Joined: 03 Sep 2010 Posts: 17 Kudos [?]: 16 [1], given: 0 Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink] ### Show Tags 24 Oct 2010, 11:30 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 [?]: 16 [1], 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, 07:09 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: 43292 Kudos [?]: 139148 [0], given: 12777 All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink] ### Show Tags 17 Sep 2017, 07:54 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, 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.
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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, 13:57
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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!(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]

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28 Oct 2017, 03:19
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!(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.

Pairing it with 20 and one twice does not make sense right? its the same. similarly to all other non -$20 bills Kudos [?]: 0 [0], given: 8 Senior Manager Joined: 02 Apr 2014 Posts: 343 Kudos [?]: 30 [0], given: 1090 Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink] ### Show Tags 04 Dec 2017, 20:20 Given: {1,1},{5,5},{10,10},{20,20} Total number ways of selecting 2 bills : 4C2 ( any 2 out of {1,5,10,20}} + 4 (one of {1,1},{5,5},{10,10},{20,20}) = 10 ways Total number of selecting two bills whose sum > 20$ = {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. Kudos [?]: 30 [0], given: 1090 Manager Joined: 17 Sep 2017 Posts: 52 Kudos [?]: 3 [0], given: 126 All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink] ### Show Tags 05 Dec 2017, 05:07 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, 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.

Where does 1/7 and 6/7 come from? Can any1 explain please

{1, 1, 5, 5, 10, 10, 20, 20} - 8 numbers

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)
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 not-20 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, 05:15
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Expert's post
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, 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.

Where does 1/7 and 6/7 come from? Can any1 explain please

{1, 1, 5, 5, 10, 10, 20, 20} - 8 numbers

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)
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 not-20 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 ! ! !

Hope 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, 05:39
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,not-20)?

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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, 05:44
lichting wrote:
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,not-20)?

No. (20, 20) is one case.
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Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2   [#permalink] 05 Dec 2017, 05:44
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