What is the probability that a 3-digit positive integer

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What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits?

A. 271/900
B. 27/100
C. 7/25
D. 1/9
E. 1/10

The OA is the option C.

Could someone give me a good explanation of how can I get the correct answer here? Please. I need some help. Thanks in advance.

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by GMATGuruNY » Tue Jul 17, 2018 1:58 am
VJesus12 wrote:What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits?

A. 271/900
B. 27/100
C. 7/25
D. 1/9
E. 1/10
P(at least one 7) = 1 - P(no 7's).

P(no 7's):
P(hundreds digit is not 7) = 8/9. (Of the 9 possible digits that could serve as the hundreds digit, eight are not 7.)
P(tens digit is not 7) = 9/10. (Of the 10 possible digits that could serve as the tens digit, nine are not 7.)
P(units digit is not 7) = 9/10. (Of the 10 possible digits that could serve as the units digit, nine are not 7.)
To combine these probabilities, we multiply:
8/9 * 9/10 * 9/10 = 18/25.

Thus:
P(at least one 7) = 1 - 18/25 = 7/25.

The correct answer is C.
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by Vincen » Tue Jul 17, 2018 2:42 am
VJesus12 wrote:What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits?

A. 271/900
B. 27/100
C. 7/25
D. 1/9
E. 1/10

The OA is the option C.

Could someone give me a good explanation of how can I get the correct answer here? Please. I need some help. Thanks in advance.
Hello Vjesus12.

This is the way I'd solve it.

We need to build a 3-digit positive number. Now,

* The first digit has 9 different options (1,2,3,...,9).
* The second digit has 10 different options (0,1,2,3,...,9).
* The third digit has 10 different options (0,1,2,3,...,9).

So, the total amount of numbers that we can build is 9*10*10=900.

On the other hand, the favorable cases are:

* One 7:
i) In the first place: there are 1*9*9=81 options.
ii) In the second place: there are 8*1*9=72 options.
ii) In the third place: there are 8*9*1=72 options.

* Two 7's:
i) One in the first place and other in the second place: there are 1*1*9=9 options.
ii) One in the first place and other in the third place: there are 1*9*1=9 options.
ii) One in the second and one in the third place: there are 8*1*1=8 options.

* Three 7's:
The only option is 777. So, 1 option.

Therefore, the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits is given by $$P=\frac{total\ favorable\ cases}{total\ options}$$ $$P=\frac{81+72+72+9+9+8+1}{900}$$ $$P=\frac{252}{900}$$ $$P=\frac{63}{225}$$ $$P=\frac{7}{25}.$$

And this is why the correct answer is the option C.

I hope it helps you.

Regards. <i class="em em-smiley"></i>

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by [email protected] » Thu Jul 19, 2018 1:47 pm
Hi All,

We're asked for the probability that a 3-digit positive integer picked at random will have one OR MORE "7s" in its digits. This question can be approached in a couple of different ways. Sometimes though, the fastest way to get to the correct answer is with a bit of 'brute force' math (here, that means 'mapping out the possible values until you recognize the inherent patterns involved).

To start, we have to determine the number of 3-digit positive integers. Those numbers are 100 to 999, inclusive. Therefore, there are 999 - 100 + 1 = 900 possible 3-digit numbers.

Next, let's look for the first few numbers that will have at least one 7 in them...Let's start with the numbers between 100 and 199:
107, 117, 127, 137, 147, 157, 167, 177, 187, 197 --> that's 10 numbers (notice that '177' has two 7s in it though...)
170, 171, 172, 173, 174, 175, 176, 177, 178, 179 --> these 10 numbers include a number that we've already 'counted' (re: 177), so we can't count it twice.
That's a total of 19 numbers between 100 and 199.

Following this same pattern, how many numbers do you think will be between 200 and 299 that have at least one 7? How about between 300 and 399? Clearly it's the same pattern, so we'll have 19 additional numbers with each group of 100... EXCEPT for the group from 700 to 799. In that group, ALL 100 numbers fit what we're looking for. Thus, we'll have 8 groups of 19 and one group of 100...

8(19) + 100 = 152+100 = 252 numbers out of 900 possible...
252/900 = 126/450 = 63/225 = 7/25

Final Answer: C

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by Scott@TargetTestPrep » Thu Jul 19, 2018 4:24 pm
VJesus12 wrote:What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits?

A. 271/900
B. 27/100
C. 7/25
D. 1/9
E. 1/10
We can do the opposite first, that is, find the probability that a random 3-digit positive integer will have no 7 in its digits. The number of such integers is:

8 x 9 x 9 = 648

(Note: There are 8 choices for the first digit (i.e., the hundreds digit) since it can't be 0 or 7. There are 9 choices for the second and third digits (i.e., the tens and units digits) since either can be any of the 10 possible digits (0 through 9, inclusive) except 7.

There are a total of 900 3-digit positive integers, 100 - 999. So the probability that a random 3-digit positive integer will have no 7 in its digits is 648/900 = 72/100 = 18/25. Therefore, the probability that a random 3-digit positive integer will have at least one 7 in its digits is 1 - 18/25 = 7/25.

Answer: C

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