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

It is currently 20 Apr 2019, 09:30

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Thurston wrote an important seven-digit phone number on a na

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

 
Manager
Manager
avatar
B
Joined: 29 Aug 2013
Posts: 74
Location: United States
Concentration: Finance, International Business
GMAT 1: 590 Q41 V29
GMAT 2: 540 Q44 V20
GPA: 3.5
WE: Programming (Computer Software)
Thurston wrote an important seven-digit phone number on a na  [#permalink]

Show Tags

New post Updated on: 12 Sep 2013, 04:31
4
16
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

46% (02:43) correct 54% (02:42) wrong based on 232 sessions

HideShow timer Statistics

Thurston wrote an important seven-digit phone number on a napkin, but the last three numbers got smudged. Thurston remembers only that the last three digits contained at least one zero and at least one non-zero integer. If Thurston dials 10 phone numbers by using the readable digits followed by 10 different random combinations of three digits, each with at least one zero and at least one non-zero integer, what is the probability that he will dial the original number correctly?

A. 1/9
B. 10/243
C. 1/27
D. 10/271
E. 1/1000000

Originally posted by shameekv on 12 Sep 2013, 04:25.
Last edited by Bunuel on 12 Sep 2013, 04:31, edited 1 time in total.
Renamed the topic and edited the question.
Intern
Intern
avatar
Joined: 03 Sep 2013
Posts: 1
Re: Thurston wrote an important seven-digit phone number on a na  [#permalink]

Show Tags

New post 12 Sep 2013, 04:33
1
3
The answer is 1/27.

Our first step is determining how many possible three-digit numbers there are with at least one zero and one nonzero. Treat this like a permutations question in which you could have any of the following six sequences, where N = non-zero integer.
0NN, N0N, NN0, N00, 00N, 0N0

There are 9 numbers that could appear in the N-slots and 1 number (zero) that could appear in the zero slots. Each sequence with two nonzero numbers will have 81 possible outcomes (1 * 9 * 9, or 9 * 1 * 9, or 9 * 9 * 1), while each sequence with one nonzero will have 9 possible outcomes (9 * 1 * 1, or 1 * 1 * 9, or 1 * 9 * 1). The total number of possible three-digit numbers here is 81 * 3 + 9 * 3 = 270.

Thurston calls 10 of these numbers, so the odds of dialing the right one are 10/270 = 1/27.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 54376
Re: Thurston wrote an important seven-digit phone number on a na  [#permalink]

Show Tags

New post 12 Sep 2013, 04:36
1
1
shameekv wrote:
Thurston wrote an important seven-digit phone number on a napkin, but the last three numbers got smudged. Thurston remembers only that the last three digits contained at least one zero and at least one non-zero integer. If Thurston dials 10 phone numbers by using the readable digits followed by 10 different random combinations of three digits, each with at least one zero and at least one non-zero integer, what is the probability that he will dial the original number correctly?

A. 1/9
B. 10/243
C. 1/27
D. 10/271
E. 1/1000000


If the last three digits have 1 zero (XX0), the total # of numbers possible is 9*9*3 (multiply by 3 since XX0 can be arranged in 3 ways: XX0, X0X, or 0XX).

If the last three digits have 2 zeros (X00), the total # of numbers possible is 9*3 (multiply by 3 since X00 can be arranged in 3 ways: X00, 00X, or X0X).

P = 10/(9*9*3+9*3) = 1/27.

Answer: C.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to the rule #3. Thank you.
_________________
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 54376
Re: Thurston wrote an important seven-digit phone number on a na  [#permalink]

Show Tags

New post 12 Sep 2013, 04:39
Bunuel wrote:
shameekv wrote:
Thurston wrote an important seven-digit phone number on a napkin, but the last three numbers got smudged. Thurston remembers only that the last three digits contained at least one zero and at least one non-zero integer. If Thurston dials 10 phone numbers by using the readable digits followed by 10 different random combinations of three digits, each with at least one zero and at least one non-zero integer, what is the probability that he will dial the original number correctly?

A. 1/9
B. 10/243
C. 1/27
D. 10/271
E. 1/1000000


If the last three digits have 1 zero (XX0), the total # of numerous possible is 9*9*3 (multiply by 3 since XX0 can be arranged in 3 ways: XX0, X0X, or 0XX).
If the last three digits have 2 zeros (X00), the total # of numerous possible is 9*3 (multiply by 3 since X00 can be arranged in 3 ways: X00, 00X, or X0X).

P=10/(9*9*3+9*3)=1/27.

Answer: C.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to the rule #3. Thank you.


Similar question to practice: john-wrote-a-phone-number-on-a-note-that-was-later-lost-94787.html
_________________
Intern
Intern
avatar
Joined: 21 Mar 2013
Posts: 38
GMAT Date: 03-20-2014
GMAT ToolKit User
Re: Thurston wrote an important seven-digit phone number on a na  [#permalink]

Show Tags

New post 11 Mar 2014, 21:09
We know that atleast one digit is Zero and atleast one digit is non-zero. The third digit can be any single digit integer (zero or non-zero).

Total # of combinations should be [One zero] * [One Non-zero] * [Any single digit integer] * \(\frac{3!}{2!}\)

= 1*9*10*3 = 270

P=10/270 = 1/27

Hence C
Director
Director
avatar
Joined: 03 Aug 2012
Posts: 695
Concentration: General Management, General Management
GMAT 1: 630 Q47 V29
GMAT 2: 680 Q50 V32
GPA: 3.7
WE: Information Technology (Investment Banking)
Re: Thurston wrote an important seven-digit phone number on a na  [#permalink]

Show Tags

New post 16 Mar 2014, 22:05
If the last three digits have 1 zero (XX0), the total # of numbers possible is 9*9*3 (multiply by 3 since XX0 can be arranged in 3 ways: XX0, X0X, or 0XX).

If the last three digits have 2 zeros (X00), the total # of numbers possible is 9*3 (multiply by 3 since X00 can be arranged in 3 ways: X00, 00X, or X0X).

P = 10/(9*9*3+9*3) = 1/27.

Answer: C.

Hi Bunuel,

Since I got this question wrong, I need insights on this.

We have two options of using either
(1).two zeros and a non-zero
or
(2). two non-zero and a zero.

In the above solution when you say XX0 can be arranged in 3 ways, since the problem is that you are considering XX as a unique single digit non-zero. However, there can be a case where 450 and 540 can be the numbers in which case the permutation will come out different.

We can consider permutations in

N00 as 3 since 0 is a unique number and we have 9 possibilities for 'N'.So, we have

9 possibilities for N and arrangement of NOO which would be !3/!2 (Divide by !2 since 0 are unique)
=27

NN0

9 possibilities for each N and arrangement of NNO which would be !3 (Not divide by !2 since N is not unique)
=9*9*6

Please suggest where I am going wrong in this one

Rgds,
TGC!
Director
Director
User avatar
Joined: 19 Apr 2013
Posts: 567
Concentration: Strategy, Healthcare
Schools: Sloan '18 (A)
GMAT 1: 730 Q48 V41
GPA: 4
GMAT ToolKit User
Re: Thurston wrote an important seven-digit phone number on a na  [#permalink]

Show Tags

New post 26 Mar 2014, 11:03
Can someone please explain why we divide 10 to 270. I know that the probability means dividing desired outcome to possible outcomes. Here desired outcome is just one number not ten.
_________________
If my post was helpful, press Kudos. If not, then just press Kudos !!!
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 54376
Re: Thurston wrote an important seven-digit phone number on a na  [#permalink]

Show Tags

New post 26 Mar 2014, 11:15
Ergenekon wrote:
Can someone please explain why we divide 10 to 270. I know that the probability means dividing desired outcome to possible outcomes. Here desired outcome is just one number not ten.


But Thurston tries 10 times not just 1:

"If Thurston dials 10 phone numbers by using the readable digits followed by 10 different random combinations of three digits, each with at least one zero and at least one non-zero integer, what is the probability that he will dial the original number correctly?"
_________________
Intern
Intern
avatar
Joined: 06 May 2013
Posts: 11
Location: United States
GMAT 1: 700 Q49 V36
Re: Thurston wrote an important seven-digit phone number on a na  [#permalink]

Show Tags

New post 30 Mar 2014, 02:38
Hi.

Please explain why after find the total possible number of the telephone numbers, we have 10 divided by 270?
I have thought that the chance that there is one correct phone numbers and 9 incorrect phone numbers is:

(1/270)*[(269/270)^9]*10!

The correct answer choice seems to indicate that each pick does not relate to the later picks, but the chance to pick the correct phone numbers increases after each pick, it isn't? That is why I multiply the chance to get correct phone numbers and the chance to get incorrect phone numbers.

What is wrong with my answer?
Intern
Intern
avatar
S
Joined: 24 Jun 2013
Posts: 32
Reviews Badge
Thurston wrote an important seven-digit phone number on a na  [#permalink]

Show Tags

New post 21 Jul 2014, 08:16
Bunuel wrote:
If the last three digits have 1 zero (XX0), the total # of numbers possible is 9*9*3 (multiply by 3 since XX0 can be arranged in 3 ways: XX0, X0X, or 0XX).

If the last three digits have 2 zeros (X00), the total # of numbers possible is 9*3 (multiply by 3 since X00 can be arranged in 3 ways: X00, 00X, or X0X).

P = 10/(9*9*3+9*3) = 1/27.

Answer: C.



Hi Bunuel,

I have a Query. In case 1 where there is only one zero, XX0 can also be XY0, in that case should it not be multiplied by 3! (i.e. 6)? For. example 3,2,0 can be written in 6 ways.

Thanks in advance for your clarification.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 54376
Re: Thurston wrote an important seven-digit phone number on a na  [#permalink]

Show Tags

New post 21 Jul 2014, 10:20
arichinna wrote:
Bunuel wrote:
If the last three digits have 1 zero (XX0), the total # of numbers possible is 9*9*3 (multiply by 3 since XX0 can be arranged in 3 ways: XX0, X0X, or 0XX).

If the last three digits have 2 zeros (X00), the total # of numbers possible is 9*3 (multiply by 3 since X00 can be arranged in 3 ways: X00, 00X, or X0X).

P = 10/(9*9*3+9*3) = 1/27.

Answer: C.



Hi Bunuel,

I have a Query. In case 1 where there is only one zero, XX0 can also be XY0, in that case should it not be multiplied by 3! (i.e. 6)? For. example 3,2,0 can be written in 6 ways.

Thanks in advance for your clarification.


The point is that 9*9 gives all possible ordered pairs of the remaining two digits:

11
12
13
14
15
16
17
18
19
21
...
99

Now, 0, in three digits can take either first, second or third place, hence multiplying by 3: XX0, X0X, 0XX.

Hope it's clear.
_________________
Manager
Manager
User avatar
Joined: 02 Jul 2012
Posts: 186
Location: India
Schools: IIMC (A)
GMAT 1: 720 Q50 V38
GPA: 2.6
WE: Information Technology (Consulting)
Reviews Badge
Thurston wrote an important seven-digit phone number on a na  [#permalink]

Show Tags

New post 15 Oct 2014, 11:01
Bunuel wrote:
arichinna wrote:
Bunuel wrote:
If the last three digits have 1 zero (XX0), the total # of numbers possible is 9*9*3 (multiply by 3 since XX0 can be arranged in 3 ways: XX0, X0X, or 0XX).

If the last three digits have 2 zeros (X00), the total # of numbers possible is 9*3 (multiply by 3 since X00 can be arranged in 3 ways: X00, 00X, or X0X).

P = 10/(9*9*3+9*3) = 1/27.

Answer: C.



Hi Bunuel,

I have a Query. In case 1 where there is only one zero, XX0 can also be XY0, in that case should it not be multiplied by 3! (i.e. 6)? For. example 3,2,0 can be written in 6 ways.

Thanks in advance for your clarification.


The point is that 9*9 gives all possible ordered pairs of the remaining two digits:

11
12
13
14
15
16
17
18
19
21
...
99

Now, 0, in three digits can take either first, second or third place, hence multiplying by 3: XX0, X0X, 0XX.

Hope it's clear.


Dear Bunuel,

I didn't get this explanation. Why are we taking XX0 and not XY0, because the non-zero numbers can also be different.
Such as

120
102
210
201
012
021

Which should lead to 6 combinations - \(3*2*1 = 6\)

Thanks
_________________
Give KUDOS if the post helps you... :-D
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 54376
Re: Thurston wrote an important seven-digit phone number on a na  [#permalink]

Show Tags

New post 15 Oct 2014, 11:16
Thoughtosphere wrote:
Bunuel wrote:
arichinna wrote:
[

Hi Bunuel,

I have a Query. In case 1 where there is only one zero, XX0 can also be XY0, in that case should it not be multiplied by 3! (i.e. 6)? For. example 3,2,0 can be written in 6 ways.

Thanks in advance for your clarification.


The point is that 9*9 gives all possible ordered pairs of the remaining two digits:

11
12
13
14
15
16
17
18
19
21
...
99

Now, 0, in three digits can take either first, second or third place, hence multiplying by 3: XX0, X0X, 0XX.

Hope it's clear.


Dear Bunuel,

I didn't get this explanation. Why are we taking XX0 and not XY0, because the non-zero numbers can also be different.
Such as

120
102
210
201
012
021

Which should lead to 6 combinations - \(3*2*1 = 6\)

Thanks


12 and 21 in your example are treated as two different numbers in my explanation. So, when I multiply by 3 I get the same result as you when you multiply by 6.

Sorry, cannot explain any better than this:
11
12
13
14
15
16
17
18
19
21
...
99

Total of 81 numbers. 0 in three digits can take either first, second or third place, hence multiplying by 3: XX0, X0X, 0XX.
_________________
Intern
Intern
avatar
B
Joined: 01 Sep 2015
Posts: 3
Re: Thurston wrote an important seven-digit phone number on a na  [#permalink]

Show Tags

New post 18 Sep 2016, 08:37
Please help me understand this -

We need to find - If Thurston dials 10 phone numbers by using the readable digits followed by 10 different random combinations of three digits, each with at least one zero and at least one non-zero integer, what is the probability that he will dial the original number correctly?.

Please consider this while counting possible outcomes. Remember, logically he will stop trying once he gets the original number. When Thurston starts dialing 10 numbers, he -
->gets the original number in 1st attempt. So he tries just 1 out of 10 number.
->gets the original number in 2nd attempt. So he tries just 2 out of 10 number.
....
...
..
gets the original number in 10th attempt. So he tries just 10 out of 10 number.


But all the explanation seems to focus on finding numbers that fit in criteria - at least one 0 and at least one non-zero for counting favorable outcomes, and not on the number that is original and ONLY ONE.

I think probability has to be calculated at two levels -

Choosing 10 numbers from all favorable outcome i.e. from 270 X (original number found at 1st attempt + original number found at 2nd attempt +......+original number found at 10th attempt).

Can somebody help where I am going wrong.
Senior Manager
Senior Manager
User avatar
G
Joined: 03 Apr 2013
Posts: 274
Location: India
Concentration: Marketing, Finance
GMAT 1: 740 Q50 V41
GPA: 3
GMAT ToolKit User
Re: Thurston wrote an important seven-digit phone number on a na  [#permalink]

Show Tags

New post 16 Jul 2017, 23:25
1
Bunuel wrote:
shameekv wrote:
Thurston wrote an important seven-digit phone number on a napkin, but the last three numbers got smudged. Thurston remembers only that the last three digits contained at least one zero and at least one non-zero integer. If Thurston dials 10 phone numbers by using the readable digits followed by 10 different random combinations of three digits, each with at least one zero and at least one non-zero integer, what is the probability that he will dial the original number correctly?

A. 1/9
B. 10/243
C. 1/27
D. 10/271
E. 1/1000000


If the last three digits have 1 zero (XX0), the total # of numbers possible is 9*9*3 (multiply by 3 since XX0 can be arranged in 3 ways: XX0, X0X, or 0XX).

If the last three digits have 2 zeros (X00), the total # of numbers possible is 9*3 (multiply by 3 since X00 can be arranged in 3 ways: X00, 00X, or X0X).

P = 10/(9*9*3+9*3) = 1/27.

Answer: C.

P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to the rule #3. Thank you.



How did you simply write 10/270?

This is how I did it.

Total possibilities for the numbers = 270 (found this one exactly how you did)

Of these only 1 is correct and the other 269 are incorrect.

Final probability = Probability of selecting 1 correct and 9 incorrect / probability of selecting any 10 out of 270

This will also give the same answer.

I just want to know your "exact mathematical logic" why you wrote 10/270.

Thank you for your help :)
_________________
Spread some love..Like = +1 Kudos :)
GMAT Club Bot
Re: Thurston wrote an important seven-digit phone number on a na   [#permalink] 16 Jul 2017, 23:25
Display posts from previous: Sort by

Thurston wrote an important seven-digit phone number on a na

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.