An integer is chosen at random between 1 and 100 find the probability that it is divisible by 8 and 2nd not divisible by 8
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Okay so, the easiest way to work this out is by listing all of the multiples of 8 from 1 to a 100.
8
16
24
32
40
48
56
64
72
80
88
96
Since the largest number is 12x8=96 we know that there are 12 possible numbers that are divisible by 8
So, the probability you choose one of these 12 numbers is 12/total possible numbers.
So, 12/100
This fraction can be simplified to 3/25
Then, 100-12 is the possibility you choose a number NOT divisible by 8.
So, 100-12=88
So the probability you choose a number NOT divisible by 8 is 88/100.
This fraction can again be simplified to 22/25.
8
16
24
32
40
48
56
64
72
80
88
96
Since the largest number is 12x8=96 we know that there are 12 possible numbers that are divisible by 8
So, the probability you choose one of these 12 numbers is 12/total possible numbers.
So, 12/100
This fraction can be simplified to 3/25
Then, 100-12 is the possibility you choose a number NOT divisible by 8.
So, 100-12=88
So the probability you choose a number NOT divisible by 8 is 88/100.
This fraction can again be simplified to 22/25.
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