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

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

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New post 22 Oct 2010, 04:52
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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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New post 22 Oct 2010, 05:35
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): \(\frac{C^2_6-1}{C^2_8}=\frac{1}{2}\).

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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New post 24 Oct 2010, 11:30
1
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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New post 17 Sep 2017, 07:09
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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New post 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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New post 21 Sep 2017, 13:57
1
1
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.

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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New post 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.

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

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New post 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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New post 05 Dec 2017, 05:15
1
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.


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

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New post 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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New post 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 &nbs [#permalink] 05 Dec 2017, 05:44
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