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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
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47% (01:52) correct 53% (01:44) wrong based on 146 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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09 Oct 2017, 02:54
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