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

by varun289 » Thu May 02, 2013 12:02 am
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?
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by Atekihcan » Thu May 02, 2013 12:10 am
Possible situations are...
  • #1. Exactly 1 zero and rest 2 non-zero ---> {0, x, x}, {x, 0, x}, and {x, x, 0} ---> 3 cases each with 9 possibilities for the 2 non-zero digits ---> 3*9*9 possibilities

    #2. Exactly 2 zeroes and rest 1 zero ---> Similar 3 cases each with 9 possibilities for the 1 non-zero digit ---> 3*9 possibilities
A total of (3*9 + 3*9*9) = 3*9*(1 + 9) = 27*10 possibilities

So, required probability = 10/(10*27) = 1/27
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by vipulgoyal » Thu May 02, 2013 9:45 pm
the probablity asks

1 zero AND 2 non zer0 OR 1 non zero AND 2 zero

1x9x9 x 6 - (3x9) + 1x1x9 x 3

why minus 3X9 :- suppose if we have 0,2,3 we could have arrange it with 3! ways

but what if it is 0,4,4 or 0,5,5 the we left with only 3-3 each options to arrenge them

so there are overlapping of 3X9 arrangements
so my ans is

10/486
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by Java_85 » Fri Sep 13, 2013 10:25 am
varun289 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?
IMO it's [spoiler]1/27[/spoiler] Here is why:

Total possible numbers: (0 10 9) & (10 0 9) & (10 9 0) --> 90+90+90 =270
We tried 10 times i.e 10 different numbers --> p=10/270=1/27
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by vipulgoyal » Sun Sep 15, 2013 11:27 pm
experts please tell me where i am wrong

1 zero and 2 non zero
case 1: when 2 non zero are differant
1*9*8 * 3! = 432
+
case 2: when 2 non zero are same
1*9*1 * 3 = 27
+
case 3: 2 zero and 1 non zero
1*1*9 * 3 = 27

required probablity = 10/486
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by theCodeToGMAT » Mon Sep 16, 2013 2:13 am
varun289 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?
2 Possible Cases
1) two 0s and one Non Zero
So, 3!(1x1x9)/2! = 27 Ways

2) two Non Zeros and one 0
Since, two digits can be same or different so take the extreme case and assume to be same:
So, 3!(1x9x9)/2!= 243 Ways
Total Ways=270.

Since, Total Ten attempts... So, 10 (1/270) = 1/27

What is the OA?
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by theCodeToGMAT » Mon Sep 16, 2013 2:58 am
vipulgoyal wrote:experts please tell me where i am wrong

1 zero and 2 non zero
case 1: when 2 non zero are differant
1*9*8 * 3! = 432
+
case 2: when 2 non zero are same
1*9*1 * 3 = 27
+
case 3: 2 zero and 1 non zero
1*1*9 * 3 = 27

required probablity = 10/486


Vipul, your application is spot-on; However, you have made a slight mistake in case 1 calculation.

case 1: when 2 non zero are differant
1*9*8 * 3! = 432

It should be "3"and not "3!".
So, 3 ( 1 x 9 x 8) = 216 + 27 + 27 = 270
Hence, 10/270 = 1/27
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by vipulgoyal » Mon Sep 16, 2013 3:55 am
my qustion remains same, in case1 all digits are differant 1 zero and two non zero differand digits, thats why i multiplied it by 3!,like the way we arrange ABC among themself
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by theCodeToGMAT » Mon Sep 16, 2013 5:48 am
vipulgoyal wrote:my qustion remains same, in case1 all digits are differant 1 zero and two non zero differand digits, thats why i multiplied it by 3!,like the way we arrange ABC among themself

Vipul, you are stuck in a same trap in which I was stuck few months ago...
For the case 1: "1 zero and 2 non zero"
According to you--> 1*9*8 * 3! = 432
BUT, consider writing same as: (1 x 9C2) x 3! = 216 (Remember we are selecting)
i.e. 9 x 8 != 1C1 x 9C2
Hope this answers your quest![/b]
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by Brent@GMATPrepNow » Mon Sep 16, 2013 5:49 am
vipulgoyal wrote:experts please tell me where i am wrong

1 zero and 2 non zero
case 1: when 2 non zero are differant
1*9*8 * 3! = 432
+
case 2: when 2 non zero are same
1*9*1 * 3 = 27
+
case 3: 2 zero and 1 non zero
1*1*9 * 3 = 27

required probablity = 10/486
To help you understand the error in your solution, I'll start by asking you to explain your rationale for 1*9*8 * 3! = 432

In particular, how did you reach 1*9*8 ?
What does the 1, the 9 and the 8 represent?

Cheers,
Brent
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by vipulgoyal » Mon Sep 16, 2013 8:51 pm
thanks Rahul and Brent,i got it, where i am wrong, actully i consider position factor twice first by 9*8 then by multipling by 3!,
Brent, by 1*9*8*3! i meant to say one zero and two non zero differant digits, now by considering 9*8 i already considered there positions (two non zero differant digits) so need to multiply by 3 non 3!
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by Brent@GMATPrepNow » Tue Sep 17, 2013 6:53 am
vipulgoyal wrote:thanks Rahul and Brent,i got it, where i am wrong, actully i consider position factor twice first by 9*8 then by multipling by 3!,
Brent, by 1*9*8*3! i meant to say one zero and two non zero differant digits, now by considering 9*8 i already considered there positions (two non zero differant digits) so need to multiply by 3 non 3!
Exactly!

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by TeonoreFossow » Wed Sep 18, 2013 3:02 am
rahulmittal87 wrote:
vipulgoyal wrote:experts please tell me where i am wrong

1 zero and 2 non zero
case 1: when 2 non zero are differant
1*9*8 * 3! = 432
+
case 2: when 2 non zero are same
1*9*1 * 3 = 27
+
case 3: 2 zero and 1 non zero
1*1*9 * 3 = 27

required probablity = 10/486


Vipul, your application is spot-on; However, you have made a slight mistake in case 1 calculation.

case 1: when 2 non zero are differant
1*9*8 * 3! = 432

It should be "3"and not "3!".
So, 3 ( 1 x 9 x 8) = 216 + 27 + 27 = 270
Hence, 10/270 = 1/27
Sometimes I really like cell phones, and change it often.Last year i got a LG Chocolate one, but at my angry moment i throw it away, so it died! And this year i bought this kind of cell phone with cheap price, but i am afraid of his fate then. I hate the noises of irrelevant calls,you know sometimes i just want to be alone, that is enough, anyone knows how can i do? I have got a cell phone jammer in Google, but I heve no idea of it, ask some profession for help!
Last edited by TeonoreFossow on Tue Sep 24, 2013 10:08 pm, edited 1 time in total.
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by theCodeToGMAT » Wed Sep 18, 2013 3:14 am
TeonoreFossow, For sure... this question can never come on GMAT :)

Please keep this forum restricted to GMAT Question discussions only.
TeonoreFossow wrote:
rahulmittal87 wrote:
vipulgoyal wrote:experts please tell me where i am wrong

1 zero and 2 non zero
case 1: when 2 non zero are differant
1*9*8 * 3! = 432
+
case 2: when 2 non zero are same
1*9*1 * 3 = 27
+
case 3: 2 zero and 1 non zero
1*1*9 * 3 = 27

required probablity = 10/486


Vipul, your application is spot-on; However, you have made a slight mistake in case 1 calculation.

case 1: when 2 non zero are differant
1*9*8 * 3! = 432

It should be "3"and not "3!".
So, 3 ( 1 x 9 x 8) = 216 + 27 + 27 = 270
Hence, 10/270 = 1/27
Sometimes I really like cell phones, and change it often.Last year i got a LG Chocolate one, but at my angry moment i throw it away, so it died! And this year i bought this kind of cell phone with cheap price, but i am afraid of his fate then. I hate the noises of irrelevant calls,you know sometimes i just want to be alone, that is enough, anyone knows how can i do? I have got a cell phone jammer in Google, but I heve no idea of it, ask some profession for help!
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by mainbhidhruv » Mon Nov 04, 2013 8:05 am
My Approach, Please let me know if this is fine....

3 digit Numbers with at least one zero and one non-zero number = 3 digit numbers - { numbers with non zero digits + number with all the digits as zero }

= 10*10*10 - {9*9*9 + 1 } = 1000 - 730 = 270

P = 10/270 = [spoiler]1/27[/spoiler]
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