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Probability Q

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by prachi18oct » Sun Jul 12, 2015 9:22 am
If each of the digits can be used, as many times as necessary, what is the probability of creating a three-digit number with three consecutive even digits in order from left to right?

A 1/80
B 1/200
C 1/360
D 1/500
E 1/720
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by theCEO » Sun Jul 12, 2015 9:50 am
prachi18oct wrote:If each of the digits can be used, as many times as necessary, what is the probability of creating a three-digit number with three consecutive even digits in order from left to right?

A 1/80
B 1/200
C 1/360
D 1/500
E 1/720
Number of 3 digit numbers = 2
2 4 6
4 6 8

Number of possibilites = 9 x 10 x 10 = 900
1st digit could be 1 to 9 = 9 choices
2nd digit could be 0 to 9 = 10 choices
2nd digit could be 0 to 9 = 10 choices

Probability = 2/900 = 1/450
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by prachi18oct » Sun Jul 12, 2015 9:52 am
theCEO wrote:
prachi18oct wrote:If each of the digits can be used, as many times as necessary, what is the probability of creating a three-digit number with three consecutive even digits in order from left to right?

A 1/80
B 1/200
C 1/360
D 1/500
E 1/720
Number of 3 digit numbers = 2
2 4 6
4 6 8

Number of possibilites = 9 x 10 x 10 = 900
1st digit could be 1 to 9 = 9 choices
2nd digit could be 0 to 9 = 10 choices
2nd digit could be 0 to 9 = 10 choices

Probability = 2/900 = 1/450
Exactly, but the options does not have this one. I guess I need to report this one.
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by theCEO » Sun Jul 12, 2015 10:17 am
prachi18oct wrote:
theCEO wrote:
prachi18oct wrote:If each of the digits can be used, as many times as necessary, what is the probability of creating a three-digit number with three consecutive even digits in order from left to right?

A 1/80
B 1/200
C 1/360
D 1/500
E 1/720
Number of 3 digit numbers = 2
2 4 6
4 6 8

Number of possibilites = 9 x 10 x 10 = 900
1st digit could be 1 to 9 = 9 choices
2nd digit could be 0 to 9 = 10 choices
2nd digit could be 0 to 9 = 10 choices

Probability = 2/900 = 1/450
Exactly, but the options does not have this one. I guess I need to report this one.
What is the source? I can bet that they wanted us to choose 1/500.
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by prachi18oct » Sun Jul 12, 2015 10:24 am
Yes the OA given is 1/500
The question is from BellCurves question Bank
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by prachi18oct » Sun Jul 12, 2015 2:40 pm
theCEO wrote:
prachi18oct wrote:
theCEO wrote:
prachi18oct wrote:If each of the digits can be used, as many times as necessary, what is the probability of creating a three-digit number with three consecutive even digits in order from left to right?

A 1/80
B 1/200
C 1/360
D 1/500
E 1/720
Number of 3 digit numbers = 2
2 4 6
4 6 8

Number of possibilites = 9 x 10 x 10 = 900
1st digit could be 1 to 9 = 9 choices
2nd digit could be 0 to 9 = 10 choices
2nd digit could be 0 to 9 = 10 choices

Probability = 2/900 = 1/450
Exactly, but the options does not have this one. I guess I need to report this one.
What is the source? I can bet that they wanted us to choose 1/500.
Why would you say so??
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by theCEO » Sun Jul 12, 2015 4:13 pm
prachi18oct wrote:
theCEO wrote:
prachi18oct wrote:
theCEO wrote:
prachi18oct wrote:If each of the digits can be used, as many times as necessary, what is the probability of creating a three-digit number with three consecutive even digits in order from left to right?

A 1/80
B 1/200
C 1/360
D 1/500
E 1/720
Number of 3 digit numbers = 2
2 4 6
4 6 8

Number of possibilites = 9 x 10 x 10 = 900
1st digit could be 1 to 9 = 9 choices
2nd digit could be 0 to 9 = 10 choices
2nd digit could be 0 to 9 = 10 choices

Probability = 2/900 = 1/450
Exactly, but the options does not have this one. I guess I need to report this one.
What is the source? I can bet that they wanted us to choose 1/500.
Why would you say so??
One could be tempted to do the following:
Number of 3 digit numbers = 2
2 4 6
4 6 8

Number of possibilites = 10 x 10 x 10 = 1000
1st digit could be 0 to 9 = 10 choices
2nd digit could be 0 to 9 = 10 choices
2nd digit could be 0 to 9 = 10 choices

Which would lead to 2/1000 = 1/500
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by MartyMurray » Sun Jul 12, 2015 8:57 pm
They need to specify either that any arrangement of 3 digits is possible including those that start with 0, in which case 1/500 would be the correct answer, or that only actual numbers be included in the set of possibilities, in which case 1/450 would be the correct answer.
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by Brent@GMATPrepNow » Mon Jul 13, 2015 6:35 am
prachi18oct wrote:If each of the digits can be used, as many times as necessary, what is the probability of creating a three-digit number with three consecutive even digits in order from left to right?

A 1/80
B 1/200
C 1/360
D 1/500
E 1/720
Some issues with this question.
First of all, the stated correct answer of 1/500 suggests that a number like 067 is considered a 3-digit number - dubious.
Second, if we were to allow for a zero in the hundreds position (e.g., 067), then we get 1000 possible 3 digit numbers, BUT we also get 3 acceptable arrangements with three consecutive even digits in order from left to right: 246, 468, AND 024
So, the probability would be 3/1000

Skip this question.

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