GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 12 Nov 2018, 20:51

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### Essential GMAT Time-Management Hacks

November 14, 2018

November 14, 2018

07:00 PM PST

08:00 PM PST

Join the webinar and learn time-management tactics that will guarantee you answer all questions, in all sections, on time. Save your spot today! Nov. 14th at 7 PM PST
• ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) # All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Manager Joined: 22 Sep 2010 Posts: 78 All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink] ### Show Tags 22 Oct 2010, 04:52 1 11 00:00 Difficulty: 95% (hard) Question Stats: 34% (02:39) correct 66% (02:47) wrong based on 148 sessions ### HideShow timer Statistics 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
Math Expert
Joined: 02 Sep 2009
Posts: 50544
Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2  [#permalink]

### Show Tags

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).

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

_________________
Intern
Joined: 03 Sep 2010
Posts: 17
Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2  [#permalink]

### Show Tags

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
Intern
Joined: 25 Aug 2017
Posts: 1
Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2  [#permalink]

### Show Tags

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.
Math Expert
Joined: 02 Sep 2009
Posts: 50544
All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2  [#permalink]

### Show Tags

17 Sep 2017, 07:54
ragini246 wrote:

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 _________________ Scott Woodbury-Stewart Founder and CEO GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Intern Joined: 03 Feb 2017 Posts: 7 Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink] ### Show Tags 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
Senior Manager
Joined: 02 Apr 2014
Posts: 471
GMAT 1: 700 Q50 V34
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.
Manager
Joined: 17 Sep 2017
Posts: 56
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. Math Expert Joined: 02 Sep 2009 Posts: 50544 Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 [#permalink] ### Show Tags 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.

22. Probability

For more:
ALL YOU NEED FOR QUANT ! ! !

Hope it helps.
_________________
Manager
Joined: 17 Sep 2017
Posts: 56
Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2  [#permalink]

### Show Tags

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)?
Math Expert
Joined: 02 Sep 2009
Posts: 50544
Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2  [#permalink]

### Show Tags

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.
_________________
Re: All envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 &nbs [#permalink] 05 Dec 2017, 05:44
Display posts from previous: Sort by