Difficult Math Problem #83 - Probability

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800guy
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Topic: Difficult Math Problem #83 - Probability
PostFri Jan 05, 2007 11:14 am
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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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thankont
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PostSat Jan 06, 2007 4:08 pm
1000/8 = 125 so prob =125/1000 = 1/8
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800guy
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PostMon Jan 08, 2007 4:47 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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anuroopa
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PostThu Mar 08, 2007 5:32 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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gabriel
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PostFri Mar 09, 2007 1:17 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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anuroopa
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PostFri Mar 09, 2007 3:17 am
yikes - bummer

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Cybermusings
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PostTue Mar 27, 2007 3:11 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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