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What is the probability that a 3digit positive integer picked at rand
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30 Dec 2009, 04:43
Question Stats:
49% (02:25) correct 51% (02:32) wrong based on 258 sessions
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What is the probability that a 3digit 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?
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:
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Re: What is the probability that a 3digit positive integer picked at rand
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30 Dec 2009, 07:31
R2I4D wrote: What is the probability that a 3digit 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? 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. 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)=1P(no 7)=1648/900=252/900=7/25 Answer: C.
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Re: What is the probability that a 3digit positive integer picked at rand
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01 Jan 2010, 10:52
Question a 3digit positive integer picked at random will have one or more "7" in its digitscan 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 3digit positive integer picked at rand
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07 Jun 2012, 22:39
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 !! 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 3digit positive integer picked at rand
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07 Jun 2012, 23:06
Bunuel wrote: R2I4D wrote: What is the probability that a 3digit 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? 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. 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)=1P(no 7)=1648/900=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 3digit positive integer picked at rand
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17 Nov 2018, 12:32
Joy111 wrote: 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 !! 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 You are considering 'only one 7' as your first case by counting numbers in the form of 7xy,7xx,7yy. What about x7y,y7x,xx7,yy7,xy7,yx7?? By counting such possible numbers, the number of events will increase. Hope this helps.



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Re: Probability of 7s
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30 Sep 2019, 02:30
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Re: Probability of 7s
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