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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?

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): \(\frac{C^2_6-1}{C^2_8}=\frac{1}{2}\).

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

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12 Sep 2017, 00:53

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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.

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)

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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