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What is the probability that a 3-digit positive integer pick

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What is the probability that a 3-digit positive integer pick

by gmatter2012 » Fri Jun 08, 2012 2:29 am
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



please note I am looking for an alternate way to solve this , to re enforce the concepts .

the way I am looking for is: At least one seven means , a three digit integer can have one 7 , two 7 or all 7 can anybody show this way ?

1- p( not all seven ) this way is understood , but I am looking for alternate approach. p( one seven ) + p( 2 seven ) + p( 3 seven )
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by Anurag@Gurome » Fri Jun 08, 2012 3:58 am
gmatter2012 wrote:...but I am looking for alternate approach. p( one seven ) + p( 2 seven ) + p( 3 seven )
That is an unnecessarily complicated approach.
Are you sure you want to go that way?

Number of 3 digit positive integers = (999 - 100) + 1 = 900

Number of 3 digit positive integers with only one 7 in it...
  • 1. Of the form 7-- : Any of the rest 9 digit can go to the rest of the two places : 9*9 = 81 integers
    2. Of the form -7- : Any of the rest 9 digit can go to the last place and any of the rest 8 integers (excluding zero) can go to the first place : 9*8 = 72 integers
    3. Of the form --7 : Any of the rest 9 digit can go to the second place and any of the rest 8 integers (excluding zero) can go to the first place : 9*8 = 72 integers

    A total of (81 + 72 + 72) = 225 integers
Number of 3 digit positive integers with only two 7 in it...
  • 1. Of the form 77- : Any of the rest 9 digit can go to the last place : 9 integers
    2. Of the form -77 : Any of the rest rest 8 integers (excluding zero) can go to the first place : 8 integers
    3. Of the form 7-7 : Any of the rest 9 digit can go to the second place : 9 integers

    A total of (9 + 8 + 9) = 26 integers
Number of 3 digit positive integers with three 7 in it...
  • Only one such integer is possible
--> P(only one 7) = 225/900
--> P(only two 7) = 26/900
--> P(three 7) = 1/900

Required probability = (225 + 26 + 1)/900 = 252/900 = 7/25

The correct answer is C.
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by Anurag@Gurome » Fri Jun 08, 2012 4:20 am
In case anyone is looking for an easier and less time consuming method...
(This is the method one should apply for solving this kind of problem not the one mentioned above)

Probability that a 3-digit positive integer picked at random will have one or more "7" in its digits = 1 - (Probability that a 3-digit positive integer picked at random will have no "7" in its digits)

Number of 3-digit positive integers = (999 - 100) + 1 = 900

For a 3-digit number with no "7" in it...
  • The first place can be filled by any of the rest 8 integers (excluding zero)
    The middle place can be filled by any of the rest 9 integers
    The last place can be filled by any of the rest 9 integers
Hence, a total of (8*9*9) = 648 such integers

Hence, required probability = 1 - 648/900 = 252/900 = 7/25

The correct answer is C.
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by gmatter2012 » Fri Jun 08, 2012 5:07 am
Thank you Anurag

I have understood your way , but it was slightly different from what I was expecting .

Here is the challenge , please take it up from here

of the form 7 _ _

1)[ seven , non seven , non seven ]

(1/9 ) * 9/10 * 9/10 * 3 = 27/100 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 as now we can have zero , from which favorable are nine, , as we are are excluding 7. And the one seven can occur in 3 ways so we multiply by 3, right?


( 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 again

( case 3) 7, 7 , 7

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


so at least one 7

27/100 + 3/100 + 1/900 = 271/900 = wrong answer !! instead of 252 /900 I am getting 271/900

what am I missing ? Please help me so that I can understand my flaw and the right concept so I do not repeat the mistakes
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by Anurag@Gurome » Fri Jun 08, 2012 5:50 am
gmatter2012 wrote:...Here is the challenge , please take it up from here

of the form 7 _ _

1)[ seven , non seven , non seven ]

(1/9 ) * 9/10 * 9/10 * 3 = 27/100 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 as now we can have zero , from which favorable are nine, , as we are are excluding 7. And the one seven can occur in 3 ways so we multiply by 3, right? [No. 7 in the first position is not same as 7 in other two positions. In those cases zero cannot go the first position. Hence, we are limited with 8 options]


( 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 again [Similar things apply here too]

( case 3) 7, 7 , 7

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


so at least one 7

27/100 + 3/100 + 1/900 = 271/900 = wrong answer !! instead of 252 /900 I am getting 271/900

what am I missing ? Please help me so that I can understand my flaw and the right concept so I do not repeat the mistakes
Following your approach of individual probability for each digit..
Only one 7
  • 1. Of the form 7-- : Any of the rest 9 digit can go to the rest of the two places : (1/9)*(9/10)*(9/10) = 9/100
    2. Of the form -7- : Any of the rest 9 digit can go to the last place and any of the rest 8 integers (excluding zero) can go to the first place : (8/9)*(1/10)*(9/10) = 8/100
    3. Of the form --7 : Any of the rest 9 digit can go to the second place and any of the rest 8 integers (excluding zero) can go to the first place : (8/9)*(9/10)*(1/10) = 8/100

    Combined probability = (9/100 + 8/100 + 8/100) = 25/100 = 1/4
Only two 7
  • 1. Of the form 77- : Any of the rest 9 digit can go to the last place : (1/9)*(1/10)*(9/10) = 1/100
    2. Of the form -77 : Any of the rest rest 8 integers (excluding zero) can go to the first place : (8/9)*(1/10)*(1/10) = 8/900 = 2/225
    3. Of the form 7-7 : Any of the rest 9 digit can go to the second place : (1/9)*(9/10)*(1/10) = 1/100 integers

    Combined probability = (1/100 + 2/225 + 1/100) = 26/900 = 13/450
Three 7
  • Only one such integer is possible : (1/9)*(1/10)*(1/10) = 1/900
Total probability = (1/4 + 13/450 + 1/900) = (225/900 + 26/900 + 1/900) = 252/900 = 7/25
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by gmatter2012 » Mon Jun 18, 2012 2:23 am
Thank you Anurag that was awesome !! This will surely help me in many similar problems , sorry for the delayed response.
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