Probability & counting problem example

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soumya029: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Mon Jul 23, 2012 1:25 pm

by soumya029 » Sun Sep 08, 2013 3:42 pm

Hi,

I was watching a video from Princeton review. It was about tackling hard maths problem. One of the probability problems explained on it has left me a little confused in relation to its answer.
the problem was "Slips of paper are numbered from 1000 to 2000.If one slip of paper is selected at random what is the probability of selecting a paper with exactly three identical digits."

The video mentioned 37/1001 as the correct answer.
I calculated a number of times and the answer came out to be 38/1001. Could anyone please explain which one of the two answers mentioned above is correct?

Thanks Soumya

Source: Beat The GMAT — Problem Solving | Sun Sep 08, 2013 3:42 pm

Java_85: Master | Next Rank: 500 Posts; Posts: 234; Joined: Tue Jul 16, 2013 9:00 am; Location: West Virginia; Thanked: 9 times

by Java_85 » Sun Sep 08, 2013 8:16 pm

1000 and 2000 2 numbers
1222 1333 1444 ... 8 possible numbers.

1011
1101
1110 We can put 23456789 instead of 0 too (But not 1) ==> 3*9=27 numbers

27+8+2=37

I also think it's 37/1001

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by [email protected] » Sun Sep 08, 2013 11:12 pm

Hi soumya029,

Go back and check your work. Did you include the number 1111 as an option? Because according to how the question is worded, you're NOT supposed to include that one.

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soumya029: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Mon Jul 23, 2012 1:25 pm

by soumya029 » Mon Sep 09, 2013 10:34 am

Yeah you are right Rich, I did include 1111. Thanks for pointing it out.

Soumya

vipulgoyal: Master | Next Rank: 500 Posts; Posts: 468; Joined: Mon Jul 25, 2011 10:20 pmFollowed by: 4 members; Thanked: 29 times

by vipulgoyal » Mon Sep 09, 2013 10:07 pm

1000,1222,1333...1999 = 9
1x11 now as of now x could be 0,2,3...9
in each case the last 3 digits can be arranged in 3!/2! ways, 3 arrangments for each value of x =
3*9 = 27 arrangements
now if x = 1, one more arrangement 1111 is possible
hence no of arrangements = 9+27+1 = 37

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