Probability

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eaakbari: Master | Next Rank: 500 Posts; Posts: 435; Joined: Mon Mar 15, 2010 6:15 amFollowed by: 1 member; Thanked: 32 times

Probability

by eaakbari » Thu Mar 25, 2010 12:45 am

In order to win a prize in the state lottery, a person must select the correct three-digit number. Jane chose 345. What is the probability that she will win? Someone told her to "box" the number to have a better chance of winning. ("Box" the number means the digits can occur in any order, such as 354,534,453, etc.) What would be the probability of Jane winning if she did "box" her numbers?

answer =[spoiler]6/1000[/spoiler]

Please explain as my asnwer is different

Last edited by eaakbari on Thu Mar 25, 2010 12:07 pm, edited 1 time in total.

Source: Beat The GMAT — Problem Solving | Thu Mar 25, 2010 12:45 am

newton9: Senior | Next Rank: 100 Posts; Posts: 71; Joined: Wed Mar 17, 2010 1:31 pmFollowed by: 1 member; Thanked: 5 times

by newton9 » Thu Mar 25, 2010 5:08 am

If she boxed the numbers,

I think the answer is 6/1000

kevincanspain: GMAT Instructor; Posts: 613; Joined: Thu Mar 22, 2007 6:17 amFollowed by: 64 members; Location: madrid; Thanked: 171 times; GMAT Score:790

by kevincanspain » Thu Mar 25, 2010 8:55 am

There are 900 3-digit numbers from 100 to 999, 6 of which are permutations of 345, so I would say 6/900= 1/150.
58 is not a 3-digit number, nor is 7

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Thu Mar 25, 2010 10:13 am

eaakbari wrote:In order to win a prize in the state lottery, a person must select the correct three-digit number. Jane chose 345. What is the probability that she will win? Someone told her to "box" the number to have a better chance of winning. ("Box" the number means the digits can occur in any order, such as 354,534,453, etc.) What would be the probability of Jane winning if she did "box" her numbers?

answer =[spoiler]1/6000[/spoiler]

Please explain as my asnwer is different

Please post your answer, all the answer choices, and the source of the question!

Kevin explained the correct answer, but without those three pieces of information it's much harder to have a useful discussion of a question.

As an aside, 1/6000 makes absolutely no sense, since there are only 900 3 digit numbers (100 to 999); so even if Jane didn't box her numbers, she'd have a 1/900 chance to win.

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eaakbari: Master | Next Rank: 500 Posts; Posts: 435; Joined: Mon Mar 15, 2010 6:15 amFollowed by: 1 member; Thanked: 32 times

by eaakbari » Thu Mar 25, 2010 12:07 pm

The question is from a high school probability book and hence has no answer choices. My answer was the same as Kevins thats is 6/900
But the book says 6/1000. Sorry for making a typo, the given answer was 6/1000 not 1/6000. Shall correct it
Thanks for the answer Kevin

nysnowboard: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Sat Mar 06, 2010 6:07 am; Thanked: 1 times

by nysnowboard » Thu Mar 25, 2010 9:58 pm

Hey all,

As it was mentioned earlier, there is definitely some ambiguity when it comes to the question itself. The earlier answer of 6/900 was appropriate for a problem that literally stated that only 3 digit INTEGERS are valid for lottery choices, but for this case, since the given answer is 6/1000, it is obvious that 007, 053, 001 ARE considered valid 3 digit LOTTERY RESULTS for this particular case.

Therefore you can approach it by saying there are 1000 three digit numbers from 000-999. For case A, her chances of winning with strictly 345, it is obviously 1/1000

For case B (boxed) as stated earlier, there are six permutations of 345, so 6/1000.

Alternatively,

There are 10 possible digits (0,1,2,3,4,5,6,7,8,9) and the chance choosing 1 of these is each 1/10

The probability of getting any in any particular order is (1/10)^3 = 1/1000

The probability of getting any permutation (which is what "boxed" really means in this case) of 3,4,5 is

3! * (1/10)^3 = 6/1000

IMHO

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