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Probability problem

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Probability problem

by topspin360 » Thu Jan 30, 2014 9:27 pm
Can someone please explain this problem in greater detail?

Why isn't the denominator 90 * 3! ( 1 * 9 * 9) * 3!?

Thanks!

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) 19
B) 10 / 243
C) 1 / 27
D) 10 / 271
E) 1 / 1000000

OA is C
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by GMATGuruNY » Thu Jan 30, 2014 10:57 pm
topspin360 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) 19
B) 10 / 243
C) 1 / 27
D) 10 / 271
E) 1 / 1000000

OA is C
Question rephrased: What is the probability that Thurston dials the last 3 digits correctly?

The last 3 digits must include at least one 0 and at least one nonzero digit.
Between 000 and 999, inclusive, there are 1000 3-digit combinations.
Since a GOOD case includes at least one 0 and at least one nonzero digit, there are two BAD cases:
Bad Case 1: all 3 digits are 0.
Bad Case 2: all 3 digits are nonzero.
All of the remaining cases will include at least one 0 and at least one nonzero digit.

Bad case 1: all 3 digits are 0
Here, there is only one option: 000.

Bad Case 2: all 3 digits are nonzero
Number of options for the 1st digit = 9. (Any digit 1-9)
Number of options for the 2nd digit = 9. (Any digit 1-9)
Number of options for the 3rd digit = 9. (Any digit 1-9)
To combine these options, we multiply:
9*9*9 = 729.

Good cases:
All of the remaining cases will include at least one 0 and at least one nonzero digit:
1000-1-729 = 270.

Of these 270 possible combinations, Thurston dials 10.
Thus, the probability that he dials the correct combination = 10/270 = 1/27.

The correct answer is C.
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by parveen110 » Tue Feb 04, 2014 6:41 am
There is another approach, by counting the # of good cases:

Case A: 2 distinct integers and 1 zero
9C2*3! = 216 ways
Case B: 2 same integers and 1 zero
9*3 = 27 ways
Case C: 2 zeros and 1 integer
9C1*3 = 27 ways

So, Total number of good cases= 216+27+27=270

Probablity=10/270=1/27
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by Bill@VeritasPrep » Tue Feb 04, 2014 10:23 am
Both good approaches. I tend to find the "total - bad cases" method to be a bit more intuitive for myself, but it's good to be comfortable with each method.
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by manhhiep2509 » Sun Mar 30, 2014 1:38 am
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.

What is wrong with my answer?
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by GMATGuruNY » Sun Mar 30, 2014 3:09 am
manhhiep2509 wrote: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.

What is wrong with my answer?
Of the 270 possible phone numbers, Thurston dials 10.

P(1st number dialed is correct) = 1/270.
P(2nd number dialed is correct) = 1/270.
P(3rd number dialed is correct) = 1/270.
P(4th number dialed is correct) = 1/270.
P(5th number dialed is correct) = 1/270.
P(6th number dialed is correct) = 1/270.
P(7th number dialed is correct) = 1/270.
P(8th number dialed is correct) = 1/270.
P(9th number dialed is correct) = 1/270.
P(10th number dialed is correct) = 1/270.

Since any of these probabilities will yield a favorable outcome, we ADD the fractions:
1/270 + 1/270 +1/270 +1/270 +1/270 +1/270 +1/270 +1/270 +1/270 +1/270 = 10/270 = 1/27.
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