Thurston wrote an important seven-digit phone number on a napkin, but the last three numbers got smudged. Thurston

[BTGmoderatorDC]

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


OA C

Source: Veritas Prep

[anonylfc]

Basically, probably of success= successful attempts/ total attempts
We are already given successful attempts, which are 10.

Now, of the 3 unreadable digits, atleast one is 0 and atleast one is non-zero.
Hence you can create the following 2 cases =
1. 0 _ _ = In the dashes there are 9 possible cases each (1,2,....,9)
Hence 9*9 = 81 cases.
But, there are 2 other variations of this arrangement _ 0 _ and _ _ 0 also. Hence total number of cases
81*3=243

2. 0 0 _ = The dashes means 9 possible cases. However, again there are 2 other variations for this arrangement 0 _ 0 or _ 0 0. Hence 9*3 = 27 cases

In all, there are 270 cases.
=> Probability = 10/27 = 1/27 (C)

[Scott@TargetTestPrep]

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


OA C

Solution:

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