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Re: If an integer is randomly chosen from the numbers 250 to 449, inclusiv [#permalink]
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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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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]
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= (\(\frac{(299-250+1)+(449-400+1)}{449-250+1}\))=\(\frac{100}{200}\)=\(\frac{1}{2}\)

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]
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]
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]
Expert Reply
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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Re: If an integer is randomly chosen from the numbers 250 to 449, inclusiv [#permalink]
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Re: If an integer is randomly chosen from the numbers 250 to 449, inclusiv [#permalink]
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