Probability Q

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prachi18oct: Master | Next Rank: 500 Posts; Posts: 269; Joined: Sun Apr 27, 2014 10:33 pmFollowed by: 5 members; Thanked: 8 times

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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Source: Beat The GMAT — Problem Solving | Sun Jul 12, 2015 9:22 am

theCEO: Master | Next Rank: 500 Posts; Posts: 363; Joined: Sun Oct 17, 2010 3:24 pmFollowed by: 3 members; Thanked: 115 times

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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prachi18oct: Master | Next Rank: 500 Posts; Posts: 269; Joined: Sun Apr 27, 2014 10:33 pmFollowed by: 5 members; Thanked: 8 times

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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theCEO: Master | Next Rank: 500 Posts; Posts: 363; Joined: Sun Oct 17, 2010 3:24 pmFollowed by: 3 members; Thanked: 115 times

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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prachi18oct: Master | Next Rank: 500 Posts; Posts: 269; Joined: Sun Apr 27, 2014 10:33 pmFollowed by: 5 members; Thanked: 8 times

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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prachi18oct: Master | Next Rank: 500 Posts; Posts: 269; Joined: Sun Apr 27, 2014 10:33 pmFollowed by: 5 members; Thanked: 8 times

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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theCEO: Master | Next Rank: 500 Posts; Posts: 363; Joined: Sun Oct 17, 2010 3:24 pmFollowed by: 3 members; Thanked: 115 times

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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MartyMurray: Legendary Member; Posts: 2135; Joined: Mon Feb 03, 2014 9:26 amFollowed by: 140 members; Location: https://martymurraycoaching.com/; Thanked: 955 times; GMAT Score:800

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.

Marty Murray
Perfect Scoring Tutor With Over a Decade of Experience
MartyMurrayCoaching.com
Contact me at [email protected] for a free consultation.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

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.

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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