) 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?
Soln: 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
Where did they get this 1000-8/8 +1 formula from? Can someone verify this answer? Thank you.
Soln: 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
Where did they get this 1000-8/8 +1 formula from? Can someone verify this answer? Thank you.
