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If an integer is randomly chosen from the numbers 250 to 449, inclusiv [#permalink]
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06 Jun 2016, 21:10
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If an integer is randomly chosen from the numbers 250 to 449, inclusive, what is the probability that all the digits of the number are even? (A) 3/5 (B) 17/50 (C) 1/8 (D) 51/199 (E) 1/4
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Last edited by DrAB on 06 Jun 2016, 22:42, edited 1 time in total.



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Re: If an integer is randomly chosen from the numbers 250 to 449, inclusiv [#permalink]
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06 Jun 2016, 21:57
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I took C as answer but it wrong...I dont know what am i missing keeping 2 as first digit , we can 6 and 8 at the second digit and 0,2,4,6,8 as third digit so ultimately , 2*5 = 10
keeping 4 at the first digit we can have 0 , 2,4 at the second, and third can be any of the 5 even. so we will have 15 total will be 25/200 = 1/8. so i guess C should be the answer.
Had it been chosen from only even then i would have think of 25/100 = 1/4



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Re: If an integer is randomly chosen from the numbers 250 to 449, inclusiv [#permalink]
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06 Jun 2016, 22:32
sudhirmadaan wrote: I took C as answer but it wrong...I dont know what am i missing keeping 2 as first digit , we can 6 and 8 at the second digit and 0,2,4,6,8 as third digit so ultimately , 2*5 = 10
keeping 4 at the first digit we can have 0 , 2,4 at the second, and third can be any of the 5 even. so we will have 15 total will be 25/200 = 1/8. so i guess C should be the answer.
Had it been chosen from only even then i would have think of 25/100 = 1/4 HI, I do not think there is anything wrong in your approach... Another way of doing it is 1)The hundreds digit can be choosen from 2,3, and 4 but number of 3s in hundreds digit is equal to total number of 2s and 4s.. so prob of picking an even number is 1/2.... 2) Probability of picking even TENS digit is 1/2 ........ 5 even from total 10 digits 3) Prbability of picking even UNITS digit is 1/2 ........ 5 even from total 10 digits.... So probability = 1/2*1/2*1/2 = 1/8 C
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Re: If an integer is randomly chosen from the numbers 250 to 449, inclusiv [#permalink]
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06 Jun 2016, 23:00
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I chose to solve this question by finding the total of numbers in the range and of numbers that fulfil the criteria, i.e. all digits are even. Total of numbers in the provided range: 449  250 +1 = 200 From 250: 1*2*5 = 10 300: none 400 to 449 = 1*3*5 = 15 Total: 12+8 = 25 Probability = \(\frac{25}{200} = \frac{5}{40} = \frac{1}{8}\)
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Re: If an integer is randomly chosen from the numbers 250 to 449, inclusiv [#permalink]
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06 Jun 2016, 23:10
DrAB wrote: If an integer is randomly chosen from the numbers 250 to 449, inclusive, what is the probability that all the digits of the number are even?
(A) 3/5 (B) 17/50 (C)1/8 (D) 51/199 (E) 1/4 Another method is to enumerate the hundreds and tens digits which are acceptable: 26 __ 28 __ 40 __ 42 __ 44 __ In each case, any of the 5 even digits (0, 2, 4, 6, 8) can take the units place. So acceptable cases = 5*5 = 25 Total number of integers = 200 Probability = 25/200 = 1/8
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Re: If an integer is randomly chosen from the numbers 250 to 449, inclusiv [#permalink]
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06 Jun 2016, 23:48
fantaisie wrote: I chose to solve this question by finding the total of numbers in the range and of numbers that fulfil the criteria, i.e. all digits are even.
Total of numbers in the provided range: 449  250 +1 = 200
From 250: 1*2*5 = 10 300: none 400 to 449 = 1*3*5 = 15
Total: 12+8 = 25
Probability = \(\frac{25}{200} = \frac{5}{40} = \frac{1}{8}\) Nice approach I did the problem on probability approach: Probability of getting even number in the UNIT digit= (\(\frac{(299250+1)+(449400+1)}{449250+1}\))=\(\frac{100}{200}\)=\(\frac{1}{2}\) Probability of f getting even number in the tens digit=\(\frac{(3+2)}{3+2+5}\)=\(\frac{5}{10}\)=\(\frac{1}{2}\) Probability of f getting even number in the Hundred digit=\(\frac{1}{2}\) So the probablity of getting Even number in all digits within this range=\(\frac{1}{2}\)*\(\frac{1}{2}\)*\(\frac{1}{2}\)=\(\frac{1}{8}\) Correct answer C
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Re: If an integer is randomly chosen from the numbers 250 to 449, inclusiv [#permalink]
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06 Jun 2016, 23:56
VeritasPrepKarishma wrote: DrAB wrote: If an integer is randomly chosen from the numbers 250 to 449, inclusive, what is the probability that all the digits of the number are even?
(A) 3/5 (B) 17/50 (C)1/8 (D) 51/199 (E) 1/4 Another method is to enumerate the hundreds and tens digits which are acceptable: 26 __ 28 __ 40 __ 42 __ 44 __ In each case, any of the 5 even digits (0, 2, 4, 6, 8) can take the units place. So acceptable cases = 5*5 = 25 Total number of integers = 200 Probability = 25/200 = 1/8 Nice approach. I didn't get your approach easily first May I know this approach more broadly ?
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If an integer is randomly chosen from the numbers 250 to 449, inclusiv [#permalink]
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07 Jun 2016, 00:17
AbdurRakib wrote: VeritasPrepKarishma wrote: DrAB wrote: If an integer is randomly chosen from the numbers 250 to 449, inclusive, what is the probability that all the digits of the number are even?
(A) 3/5 (B) 17/50 (C)1/8 (D) 51/199 (E) 1/4 Another method is to enumerate the hundreds and tens digits which are acceptable: 26 __ 28 __ 40 __ 42 __ 44 __ In each case, any of the 5 even digits (0, 2, 4, 6, 8) can take the units place. So acceptable cases = 5*5 = 25 Total number of integers = 200 Probability = 25/200 = 1/8 Nice approach. I didn't get your approach easily first May I know this approach more broadly ? I think this is the same approach as the one I presented above, just more of a shortcut. She first evaluates how many different combinations of the two first digits are possible, thus the list of 2 digit combinations (hundrends and tenths) which in total equals to 5 different combinations that satisfy the requirements given by the range and also that all digits are even. She then checks how many different numbers would be applicable to each combination of digits, which is also five. So every 2 digit combination will have 5 options for the last digit, eg: 260, 262, 264, 266, 268 280, 282, 284, 286, 288 I hope it helps!
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Re: If an integer is randomly chosen from the numbers 250 to 449, inclusiv [#permalink]
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07 Jun 2016, 22:54
AbdurRakib wrote: VeritasPrepKarishma wrote: DrAB wrote: If an integer is randomly chosen from the numbers 250 to 449, inclusive, what is the probability that all the digits of the number are even?
(A) 3/5 (B) 17/50 (C)1/8 (D) 51/199 (E) 1/4 Another method is to enumerate the hundreds and tens digits which are acceptable: 26 __ 28 __ 40 __ 42 __ 44 __ In each case, any of the 5 even digits (0, 2, 4, 6, 8) can take the units place. So acceptable cases = 5*5 = 25 Total number of integers = 200 Probability = 25/200 = 1/8 Nice approach. I didn't get your approach easily first May I know this approach more broadly ? Here is the way I thought about it: The number has to be chosen from 250 to 449. All digits have to be even. Let's consider the acceptable hundreds and tens digits : 26 __  The units digit can be any even digit since all numbers in the range 260 to 269 are included. 28 __  The units digit can be any even digit since all numbers in the range 280 to 289 are included. 3 __ __  All such numbers are not acceptable since 3 is odd. Move on to 4 __ __ 40 __  The units digit can be any even digit since all numbers in the range 400 to 409 are included. 42 __  The units digit can be any even digit since all numbers in the range 420 to 429 are included. 44 __  The units digit can be any even digit since all numbers in the range 440 to 449 are included. Each of the acceptable 5 options have further 5 options for the units digit: 260, 262, 264, 266, 268 280, 282, 284, 286, 288 ... and so on So there will be 25 such numbers.
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