SOLUTION: What is the probability that a number selected at random from the first 500 positive integers is (exactly) divisible by 6 or 9?

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Question 346172: What is the probability that a number selected at random from the first 500 positive integers is (exactly) divisible by 6 or 9?
Found 2 solutions by Fombitz, CharlesG2:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
There are 83 numbers exactly divisible by 6 from 6 to 498.
There are 55 numbers exactly divisible by 9 from 9 to 495.
There are 27 numbers that are exactly divisible by 6 and 9 from 18 to 486, we don't want to double count those.
There are 500 numbers total.
P=83%2F500%2B55%2F500-27%2F500
P=111%2F500

Answer by CharlesG2(834) About Me  (Show Source):
You can put this solution on YOUR website!
What is the probability that a number selected at random from the first 500 positive integers is (exactly) divisible by 6 or 9?


using as the first 500 positive integers 1 through 500 inclusively


there are 83 numbers divisible by 6 exactly between 1 and 500
(83 * 6 = 498)
there are 55 numbers divisible by 9 exactly between 1 and 500
(55 * 9 = 495)

probability of 6: 83/500 = 16.6/100 = 0.166 = 16.6 %

probability of 9: 55/500 = 11/100 = 11 %

probability of number being divisible by either 6 or 9: (83 + 55)/500
138/500 = 27.6/100 = 27.6 %

probability of number being divisible by either 6 or 9 but not both:

there are 27 numbers divisible by both 6 and 9 exactly between 1 and 500
(27 * 18 = 486)

138 - 27 = 111

probability of number being divisible by either 6 or 9 but not both:
111/500 = 22.2/100 = 22.2 %