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

Question

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

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

apurvt · Accepted Answer

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.

Bunuel · Answer

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.

Bunuel · Answer

Similar question to practice: john-wrote-a-phone-number-on-a-note-that-was-later-lost-94787.html

idinuv · Answer

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   *   *   *  \frac{3!}{2!}

= 1*9*10*3 = 270

P=10/270 = 1/27

Hence C

TGC · Answer

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!

Ergenekon · Answer

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.

Bunuel · Answer

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?"

phammanhhiep · Answer

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)* *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?

arichinna · Answer

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.

Bunuel · Answer

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.

Thoughtosphere · Answer

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

Bunuel · Answer

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.

BrajeshB · Answer

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.

ShashankDave · Answer

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

kitzi · Answer

The combination of the last 3 numbers could be 0ZZ (where I assume Z is the non-zero integer), which will be 3C1 x 81, or 00Z, which will be 3x9. In total we will have 270 combinations. It is indeed that out of that 270 combinations available there is only 1 correct combination, in which the probability is 1/270. However, given that he attempted 10 phone calls, it is possible that he will get the correct combination either in the 1st attempt, 2nd attempt.... or even in the 10th attempt. Each attempt has a probability of 1/270. So the total probability of him getting the right combination in 10 tries is 10x1/270 = 1/27.

ScottTargetTestPrep · Answer

We can divide the last 3 digits of the phone number into 2 cases:

1) exactly 1 zero and 2 non-zero digits.

2) exactly 2 zeros and 1 non-zero digit.

Case 1: ZNN, NZN, NNZ (where Z is the 0 digit and N is a nonzero digit)

(1 x 9 x 9) x 3 = 243

Case 2: ZZN, ZNZ, NZZ

(1 x 1 x 9) x 3 = 27

Therefore, the total number of ways the last 3 digits of the phone number can be formed given that there is at least one zero and at least one non-zero digit is 243 + 27 = 270. Since Thurston tries 10 of them, the probability he dials the correct phone number is 10/27 = 1/27.

Answer: C

Wadree · Answer

At least one 0   .....and at least one nonzero   ..... we gotta take......and we got 3 spots to fill.....
We can possibly take 1 zero.....or 2 zero.......we cant take 3 zero because then no spot will be left for at least 1 nonzero.......

Now there are 2 possible scenario..... 
1. We take 1 zero and the other 2 spots will be automatically non zero
2. We take 2 zero and the other 1 spot will be nonzero....think of this scenario as we take 1 nonzero and other 2 spots will be automatically zero......

So..... when we take 1 zero.....

First sub scenario : 1st spot is zero.....so second and third spot can each have 9 possible digits.....so 9 × 9 = 81 possibility....

Second sub scenario : 2nd spot is zero.....so first and third spot can each have 9 possible digits....so 9 × 9 = 81 possibility....

Third sub scenario : 3rd spot is zero.....so first and second spot can each have 9 possible digits....so 9 × 9 = 81 possibility....

And when we take 1 nonzero.....

First sub scenario : first spot is nonzero..... So 9 possiblity

Second sub scenario : second spot is nonzero..... So 9 possiblity

Third sub scenario : third spot is nonzero.... So 9 possiblity

So..... Total scenario = 81+81+81+9+9+9 = 270

And 1 scenario is real one among these..... So probability should be 1 / 270.....buuut......he will try 10 scenarios.....so his probability is 10 ×   = 1 / 27 ........

!nah id win!

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

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