Difficult Math Problem #83 - Probability

1000/8 = 125 so prob =125/1000 = 1/8

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

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

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

yikes - bummer

thanks

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.