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What is the probability that a 3-digit positive integer picked at rand

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What is the probability that a 3-digit positive integer picked at rand [#permalink]

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New post 30 Dec 2009, 03:43
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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

[Reveal] Spoiler:
I guessed A, pressed for time. How would I do this problem without manually figuring out how many digits have at least one 7?


When I did it again for correctness, I did the following:

|{700...799}| = 100 integers.
|{107, 117, 127, 137, 147, 157, 167, 187, 197}| = 9 per set of 100, therefore, 72 integers (8 sets, excluding 700s).
|{170, 171, 172, 173, 174, 175, 176, 178, 179}| = 9 per set of 100, therefore, 72 integers (8 sets, excluding 700s).
|{177, 277, 377, 477, 577, 677, 877, 977}| = 8 integers

Total integers with at least one 7 = 100 + 72 + 72 + 8 = 252
Total outcomes = 999 - 100 + 1 = 900

Probability = 252/900 = 7/25. That. took. forever. :evil:
[Reveal] Spoiler: OA

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Re: What is the probability that a 3-digit positive integer picked at rand [#permalink]

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New post 30 Dec 2009, 06:31
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R2I4D 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

I guessed A, pressed for time. How would I do this problem without manually figuring out how many digits have at least one 7?

[Reveal] Spoiler:
OA is C.


When I did it again for correctness, I did the following:

|{700...799}| = 100 integers.
|{107, 117, 127, 137, 147, 157, 167, 187, 197}| = 9 per set of 100, therefore, 72 integers (8 sets, excluding 700s).
|{170, 171, 172, 173, 174, 175, 176, 178, 179}| = 9 per set of 100, therefore, 72 integers (8 sets, excluding 700s).
|{177, 277, 377, 477, 577, 677, 877, 977}| = 8 integers

Total integers with at least one 7 = 100 + 72 + 72 + 8 = 252
Total outcomes = 999 - 100 + 1 = 900

Probability = 252/900 = 7/25. That. took. forever. :evil:


There are total 900 3 digit numbers;

3 digit number with no 7 =8*9*9=648 (first digit can take 8 values from 1 to 9 excluding 7; second and third digits can take 9 values from 0 to 9 excluding 7),

P(at least one 7)=1-P(no 7)=1-648/900=252/900=7/25

Answer: C.
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Re: What is the probability that a 3-digit positive integer picked at rand [#permalink]

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New post 01 Jan 2010, 09:52
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Question a 3-digit positive integer picked at random will have one or more "7" in its digits
can be approached by finding out the number of positive integers which would not have 7 at all and then subtracting the posb from the overall posb combinations.

3 Digit Number posb = 9 x 10 x 10 = 900

3 digit number not having 7 in it all = 8 x 9 x 9 = 648

Therefore 3 digits number having atleast one 7 in them = 900 - 648 = 252

Hence probability \(= \frac{252}{900} = \frac{7}{25}\). Hence answer is C
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Re: What is the probability that a 3-digit positive integer picked at rand [#permalink]

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New post 07 Jun 2012, 21:39
some one please point the flaw in my logic

want to do it the longer way for more understanding .

at least one 7 means , one , two or three 7's

lets take scenarios, first digit 7 , two digits 7 , all three digits 7

1) first digit to be seven , non seven , non seven

(1/9 ) * 9/10 * 9/10 = 81/900=9/100 first digit cannot be zero , so there are 9 total possibilities for the first , from

1 ,2 , 3, 4, 5, 6, 7, 8, 9. For the second and third we have 10 possibilities now we can have zero , from which favorable are nine, , as we are are excluding 7


( case 2) 7 , 7 , non seven

1/9 * 1/10 *9/10 * 3 = 3/100 here we are multiplying by 3 , because the 2 seven's can occur in 3 ways

( case 3) 7, 7 , 7

1/9 * 1/10 * 1/10 = 1/ 900


so at least one 7

9/100 + 3/100 + 1/900 = 109/900 = wrong answer !! :oops:

what am I missing guys ?

I know the alternate way , 1- p ( all not seven ) , that's fine , but I want to understand what's wrong with this way .

Appreciate your help. Thank you

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Re: What is the probability that a 3-digit positive integer picked at rand [#permalink]

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New post 07 Jun 2012, 22:06
Bunuel wrote:
R2I4D 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

I guessed A, pressed for time. How would I do this problem without manually figuring out how many digits have at least one 7?

[Reveal] Spoiler:
OA is C.


When I did it again for correctness, I did the following:

|{700...799}| = 100 integers.
|{107, 117, 127, 137, 147, 157, 167, 187, 197}| = 9 per set of 100, therefore, 72 integers (8 sets, excluding 700s).
|{170, 171, 172, 173, 174, 175, 176, 178, 179}| = 9 per set of 100, therefore, 72 integers (8 sets, excluding 700s).
|{177, 277, 377, 477, 577, 677, 877, 977}| = 8 integers

Total integers with at least one 7 = 100 + 72 + 72 + 8 = 252
Total outcomes = 999 - 100 + 1 = 900

Probability = 252/900 = 7/25. That. took. forever. :evil:


There are total 900 3 digit numbers;

3 digit number with no 7 =8*9*9=648 (first digit can take 8 values from 1 to 9 excluding 7; second and third digits can take 9 values from 0 to 9 excluding 7),

P(at least one 7)=1-P(no 7)=1-648/900=252/900=7/25

Answer: C.


Oh man my mistake was 9 * 9 * 9 and then you mentioned that first digit can only take values from 1 2 3 4 5 6 7 8 9 and not 0 because that will make it a 2 digit number.

Argh! traps traps traps!
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Re: What is the probability that a 3-digit positive integer picked at rand [#permalink]

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Re: What is the probability that a 3-digit positive integer picked at rand   [#permalink] 09 Oct 2017, 01:54
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