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An integer x, with 10 <= x <= 99, is to be chosen. If all choices are
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21 Mar 2019, 23:10
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58% (01:42) correct 42% (01:33) wrong based on 45 sessions
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An integer x, with \(10\leq x\leq 99\), is to be chosen. If all choices are equally likely, what is the probability that at least one digit of x is a 7? (A) 1/9 (B) 1/5 (C) 19/90 (D) 2/9 (E) 1/3
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An integer x, with 10 <= x <= 99, is to be chosen. If all choices are
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Updated on: 23 Mar 2019, 00:35
Bunuel wrote: An integer x, with \(10\leq x\leq 99\), is to be chosen. If all choices are equally likely, what is the probability that at least one digit of x is a 7?
(A) 1/9 (B) 1/5 (C) 19/90 (D) 2/9 (E) 1/3 total integerts = 9910 +1 ; 90 so with 7 ; 19 integers possible from 10 to 99 P = 19/90 IMO C @bunuel; is given answer correct? if yes then please provide solution
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Originally posted by Archit3110 on 22 Mar 2019, 01:50.
Last edited by Archit3110 on 23 Mar 2019, 00:35, edited 1 time in total.



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Re: An integer x, with 10 <= x <= 99, is to be chosen. If all choices are
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22 Mar 2019, 21:58
Bunuel wrote: An integer x, with \(10\leq x\leq 99\), is to be chosen. If all choices are equally likely, what is the probability that at least one digit of x is a 7?
(A) 1/9 (B) 1/5 (C) 19/90 (D) 2/9 (E) 1/3 I got answer 1/5. But i am not sure how it is right. Request Bunnel to explain.. Thanks



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Re: An integer x, with 10 <= x <= 99, is to be chosen. If all choices are
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22 Mar 2019, 22:45
Bunuel wrote: An integer x, with \(10\leq x\leq 99\), is to be chosen. If all choices are equally likely, what is the probability that at least one digit of x is a 7?
(A) 1/9 (B) 1/5 (C) 19/90 (D) 2/9 (E) 1/3 optn B... 18/90..... OA pls... Sent from my iPhone using GMAT Club Forum



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Re: An integer x, with 10 <= x <= 99, is to be chosen. If all choices are
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23 Mar 2019, 00:36
Abhigmat2019 wrote: Bunuel wrote: An integer x, with \(10\leq x\leq 99\), is to be chosen. If all choices are equally likely, what is the probability that at least one digit of x is a 7?
(A) 1/9 (B) 1/5 (C) 19/90 (D) 2/9 (E) 1/3 optn B... 18/90..... OA pls... Sent from my iPhone using GMAT Club Forum Abhigmat201977 has 2 7's so it will be 19 7s...
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Re: An integer x, with 10 <= x <= 99, is to be chosen. If all choices are
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23 Mar 2019, 00:37
SWAPNILP wrote: Bunuel wrote: An integer x, with \(10\leq x\leq 99\), is to be chosen. If all choices are equally likely, what is the probability that at least one digit of x is a 7?
(A) 1/9 (B) 1/5 (C) 19/90 (D) 2/9 (E) 1/3 I got answer 1/5. But i am not sure how it is right. Request Bunnel to explain.. Thanks SWAPNILPplease share your solution of how you got 1/5.
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Re: An integer x, with 10 <= x <= 99, is to be chosen. If all choices are
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23 Mar 2019, 01:30
I initially picked C, but the problem is we are taking '77' twice while getting to 19. Therefore, it should be 18/90 = 1/5
Option B



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Re: An integer x, with 10 <= x <= 99, is to be chosen. If all choices are
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24 Mar 2019, 02:18
Archit3110 wrote: SWAPNILP wrote: Bunuel wrote: An integer x, with \(10\leq x\leq 99\), is to be chosen. If all choices are equally likely, what is the probability that at least one digit of x is a 7?
(A) 1/9 (B) 1/5 (C) 19/90 (D) 2/9 (E) 1/3 I got answer 1/5. But i am not sure how it is right. Request Bunnel to explain.. Thanks SWAPNILPplease share your solution of how you got 1/5. Hi Bunuel Please find below my approach: Step 1: total no of integers between 10 and 99 = 90 Step 2: Integers with digit 7 = 18 Step 3: Probability = 18C1/90C1 = 18/90 = 1/5 Please advise.



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Re: An integer x, with 10 <= x <= 99, is to be chosen. If all choices are
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24 Mar 2019, 03:16
Bunuel wrote: An integer x, with \(10\leq x\leq 99\), is to be chosen. If all choices are equally likely, what is the probability that at least one digit of x is a 7?
(A) 1/9 (B) 1/5 (C) 19/90 (D) 2/9 (E) 1/3 We have 9910 = 89+1 (inclusive set) = 90 integers total.. We are asked AT LEAST ONE DIGIT OF X TO BE A 7 i.e we are okay even if we get a 77 and it won't be double counting because "At least one 7 means", 2 or 3 or 4 or 5 or , as many 7's as possible are welcome in this combination.. but since the upper limit is 99, we have to stop at 97 here, because that's the last digit with AT LEAST ONE 7 In it. therefore, we start with first digit with a 7 in it and go on... 17, 27, 37, 47, 57, 67, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 87, 97 .. We get 18 integers with AT least one 7.. and when divided by 90, which is the total number of integers in the given list. we get 18/90 = 1/5 (B) as the answer.



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Re: An integer x, with 10 <= x <= 99, is to be chosen. If all choices are
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24 Mar 2019, 03:47
Bunuel wrote: An integer x, with \(10\leq x\leq 99\), is to be chosen. If all choices are equally likely, what is the probability that at least one digit of x is a 7?
(A) 1/9 (B) 1/5 (C) 19/90 (D) 2/9 (E) 1/3 HOW TO FIND NUMBER OF TERMS FROM A TO Z \(\frac{last..term  first..term}{2} +1\) First lets find number of terms between 10 and 99 inclusive \(\frac{99  10}{2} +1\) = \(45\) we can have following numbers with at least one digit 7. And why everyone loves this number 7 17, 27, 37, 47, 57, 67, 77, 87, 97 So total 9 numbers Hence the probability that at least one digit of x is a 7 = \(\frac{9}{45}\) or \(\frac{1}{5}\)



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Re: An integer x, with 10 <= x <= 99, is to be chosen. If all choices are
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24 Mar 2019, 03:52
dave13any reason why you havent considered no from 70 to 79...? dave13 wrote: Bunuel wrote: An integer x, with \(10\leq x\leq 99\), is to be chosen. If all choices are equally likely, what is the probability that at least one digit of x is a 7?
(A) 1/9 (B) 1/5 (C) 19/90 (D) 2/9 (E) 1/3 HOW TO FIND NUMBER OF TERMS FROM A TO Z \(\frac{last..term  first..term}{2} +1\) First lets find number of terms between 10 and 99 inclusive \(\frac{99  10}{2} +1\) = \(45\) we can have following numbers with at least one digit 7. And why everyone loves this number 7 17, 27, 37, 47, 57, 67, 77, 87, 97 So total 9 numbers Hence the probability that at least one digit of x is a 7 = \(\frac{9}{45}\) or \(\frac{1}{5}\)
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An integer x, with 10 <= x <= 99, is to be chosen. If all choices are
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24 Mar 2019, 04:00
Archit3110 wrote: dave13any reason why you havent considered no from 70 to 79...? Archit3110i actually used a shortcut, cause its less time consuming, well i think so i used these numbers: 17, 27, 37, 47, 57, 67, 77, 87, 97 but if i reverse these numbers i get: 71, 72, 73, 74, 75, 76, 78, 79 and also 70 So total numbers now if you see 18: \(\frac{18}{90}\) = \(\frac{1}{5}\) same result So if I consider, 70 to 79, then I have to consider the rest numbers that I have reversed



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Re: An integer x, with 10 <= x <= 99, is to be chosen. If all choices are
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24 Mar 2019, 04:07
dave13 wrote: Archit3110 wrote: dave13any reason why you havent considered no from 70 to 79...? Archit3110i actually used a shortcut, cause its less time consuming, well i think so i used these numbers: 17, 27, 37, 47, 57, 67, 77, 87, 97 but if i reverse these numbers i get: 71, 72, 73, 74, 75, 76, 78, 79 and also 70 So total numbers now if you see 18: \(\frac{18}{90}\) = \(\frac{1}{5}\) same result dave13ok, understood, but i dnt think this trick would be always helpful.. anyways thanks for enlightening on an alternate method..
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Re: An integer x, with 10 <= x <= 99, is to be chosen. If all choices are
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06 Apr 2019, 12:00
If anyone's interested in a solution without counting all possible number, here's one
Number of ways a number can be picked = (9910) + 1=> 90 An easier way to calculate probability to pick a number with atleast one 7 is: 1  (Probability of picking a number with no 7s)
Probability of picking a number with no 7s Since the number has to be a two digit number, number of ways in which its ten's digit can be selected is 8 (since we can't use 7 or 0) AND, number of ways in which its ten's digit can be selected is 9 (since we can't use 7) Therefore, Probability of picking a number with no 7s = 72/90 or 4/5
And hence probability to pick a number with atleast one 7= 1  4/5 = 1/5




Re: An integer x, with 10 <= x <= 99, is to be chosen. If all choices are
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06 Apr 2019, 12:00






