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Difficult Math Problem #83 - Probability

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by thankont » Sat Jan 06, 2007 3:08 pm
1000/8 = 125 so prob =125/1000 = 1/8
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OA

by 800guy » Mon Jan 08, 2007 3:47 pm
OA:

Any multiple of 8 is also a multiple of 2 so we need to find the multiples of 8 from 0 to 1000
the first one is 8 and the last one is 1000
==> ((1000-8 )/8 ) + 1 = 125
==> p (picking a multiple of 2 & 8 ) = 125/1000 = 1/8
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help

by anuroopa » Thu Mar 08, 2007 4:32 am
can 1 of u help me on this - since the q asks for the first 1000 positive integers- does it it not mean 0 to 9999 and hence the number of multiples of 8 should be 124 . shouldn't the probability then be 124 / 1000
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Re: help

by gabriel » Fri Mar 09, 2007 12:17 am
anuroopa wrote:can 1 of u help me on this - since the q asks for the first 1000 positive integers- does it it not mean 0 to 9999 and hence the number of multiples of 8 should be 124 . shouldn't the probability then be 124 / 1000
0 is neither positive nor negative ...... so the first 1000 positive integers wuld be from 1 to 1000 and hence the probability is 125/1000=1/8
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by anuroopa » Fri Mar 09, 2007 2:17 am
yikes - bummer

thanks
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by Cybermusings » Tue Mar 27, 2007 2:11 am
It's a pretty easy one...Just make it look like a tough one to crack....

Any number divisible by 8 is by default divisible by 2 (Remember 2 is a multiple of 8!!)

Hence the sequence be like 8,16,24,32,40,48....1000

Now 1000/8=125 (Hence 1000 is the 125th term of the sequence)

Now the probability = 125/1000
= 1/8

Hope this Helps

Cheers,

Rashi.
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