Problem Solving

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?

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.