**Quote:**

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

Begin by thinking about the problem logically to most efficiently address the problem, while simultaneously eliminating any choices that are proven logically impossible.

Based on the scenario, there is only one correct phone number, so logically eliminate Choices B and D for having a value other than one in the numerator.

Then, recognize that were there no limitations on the final three digits, there would be 10 x 10 x 10 = 10,000 possible combinations. Obviously, the restrictions from the problem will result in there being fewer possibilities, so also logically eliminate Choice E for being much greater than 10,000.

Now, logically, we could also recognize that one of the three digits is unrestricted and could be any of the ten digits, therefore it wouldn't make sense for there to be a larger than 1/10 chance of guessing the correct number as in Choice A. Through logical estimation alone it is reasonable to potentially arrive to the correct answer of Choice C without needing to complete the complex set up to specifically solve for this problem.

Still, lets consider the ways to accomplish the restrictions listed by the problem. First, recognize that order matters, so consider the 0 as the control digit.

If the 0 is in the first digit, then there are 90 possible permutations because one of the other digits must be a non-zero digit and there would be 9 x 10 possibilities.

This 90 possible permutations would remain the same were the 0 in the second or third digits as well.

Since the 0 could be in the first or the second or the third, add those three sets of 90 permutations together to a total of 270 possible permutations.

Finally, remember that there were ten attempts so that probability of 10/270 reduces to 1/27 which matches Choice C.

_________________

Stefan Maisnier

Director of Online Tutoring, MyGuru

www.myguruedge.com

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