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Thurston wrote an important seven-digit phone number on a na [#permalink]

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12 Sep 2013, 04:25

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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?

Re: Thurston wrote an important seven-digit phone number on a na [#permalink]

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12 Sep 2013, 04:33

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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.

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).

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).

Re: Thurston wrote an important seven-digit phone number on a na [#permalink]

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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!
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Re: Thurston wrote an important seven-digit phone number on a na [#permalink]

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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.
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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?"
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Re: Thurston wrote an important seven-digit phone number on a na [#permalink]

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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.

Thurston wrote an important seven-digit phone number on a na [#permalink]

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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.

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.

Thurston wrote an important seven-digit phone number on a na [#permalink]

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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.

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.

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.
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Re: Thurston wrote an important seven-digit phone number on a na [#permalink]

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29 Apr 2016, 08:54

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Re: Thurston wrote an important seven-digit phone number on a na [#permalink]

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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).

Re: Thurston wrote an important seven-digit phone number on a na [#permalink]

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16 Jul 2017, 23:25

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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).