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Difficult Math Problem #83 - Probability

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by 800guy » Fri Jan 05, 2007 10:14 am
answer coming after some people try answering..

If one number is chosen at random from the first 1000 positive integers, what is the probability that the number chosen is multiple of both 2 and 8?
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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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