Pick a random number from 1 to 1000 Find the probability tha

Pick a random number from 1 to 1000. Find the probability that this number is not divisible by any of the numbers 2, 3, 5.

Solution

Let the set C be the set of all the numbers. So C = {1,2,3,...1000}
Number of elements of C = cardinality of C = card(C) = 1000.
Now for finding out the numbers not divisible by 2,3 or 5, we should focus on the numbers divisible of 2,3 and 5.
Let the set of numbers divisible by 2 be C2.
So C2= {2,4,6...998,1000} = {2x1, 2x2, 2x3 ... 2x449, 2x500} = 2 x {1,2,3 ... 449, 500}
Its cardinality is card(C2) = 500.

Let the set of numbers divisible by 3 be C3.
So C3= {3,6,9... 996,999} = {3x1, 3x2, 3x3... 3x332, 3x333} = 3 x {1,2,3 ... 332, 333}
Its cardinality is card(C3)= 333.

Let the set of numbers divisible by 5 be C5.
So C5 = {5,10,15... 995,1000} = {5x1, 5x2, 5x3... 5x199, 5x200} = 5 x {1,2,3...199,200}
Its cardinality is card(C5) = 200.

Now we need to consider the numbers which are divisible by both(2,3), (3,5) and (2,5) along with the numbers divisible by (2,3,5)

Let the set of numbers divisible by (2 and 3) 6 be C2x3 = C6
C6 = {6,12,18,...996} = 6x{1,2... 166}
Notice that these numbers are already taken care of in C2 and C3.
card(C6) = 166.

Similarly C10 = set of numbers divisible by 10 (2,5) = {10,20,30...990,1000}= 10 x {1,2,3...99,100}
These are in C2 and C5.
card(C10) = 100.

Similarly C15 = set of numbers divisible by 15 (3,5) = {15,30,45...990}= 15 x {1,2,3...66}
These are in C3 and C5.
card(C15) = 66.

Now the set C30 be the set which contains the numbers divisible by 2,3 and 5.
C30 = {30,60,90,... 990} = 30x {1,2,3.. 33}
These numbers are twice ignored in C6,C10 and C15.
card(C30)= 33

Now the numbers divisible by any of the digits 2 , 3 or 5 is card(C2) + card(C3) + card(C5) - card(C6) - card(C10) - card(15) + card(C30) = 500+333+200-166-100-66+33 = 734

So the numbers within 1000 not divisible by 2 3 or 5 is 1000-734 = 266
Answer is the unions of Set C2 , C3 and C5.