tzec76 wrote:

I am not sure about this. Can someone assist?

If one number is chosen at random from the first 1,000

positive integers, what is the probability that the number

chosen is a multiple of both 2 and 8?

a. 1/125

b. 1/8

c. 1/2

d. 9/16

e. 5/8

first find how many numbers are multiples of both 2 and 8

simply put, every number divisible by 8 will be divisible by 2

first # divisible by 8 .. 8

last # divisible by 8 ...1000

total #'s ......(1000-8)/8 +1 = 124 +1= 125

total numbers = 1000

prob = 125 /1000 = 1/8