Question 1132434: If choosing two numbers between 25 and 225, what is the probability of choosing two numbers that are divisible by 4 but not divisible by either 6 or 10?
Found 2 solutions by greenestamps, Boreal:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
The least common multiple of 4, 6, and 10 is 60. So the pattern of which multiples of 4 are or are not multiples of either 6 or 10 will be a cycle of length 60.
The given interval, 25 to 225, is an interval of length 200.
Since the given interval is not an integer multiple of the cycle length, we need to be careful in our counting.
In each cycle of length 60, there are 15 numbers divisible by 4, 5 divisible by 12 (LCM of 4 and 6), 3 divisible by 20 (LCM of 4 and 10), and 1 divisible by 60 (LCM of 4, 6, and 10). So in each cycle of length 60, the number of integers divisible by 4 but not by 6 or 10 is (15 - (5+3) + 1) = 8.
So in each interval of length 60, there are 8 integers that are multiples of 4 that are not divisible by either 6 or 10, and there are 15-8 = 7 integers that are multiples of 4 and ARE divisible by either 6 or 10.
The complete interval of 200 is 3 cycles of 60 (28-84, 88-144, and 148-204), plus a partial cycle of 20 (208-224).
3 cycles of 60 gives us 24 integers divisible by 4 and not divisible by either 6 or 10 and 21 integers divisible by 4 but also divisible by either 6 or 10.
Then we need to look at the multiples of 4 from 208 to 224. 2 of them are divisible by either 6 or 10; the other 3 are not.
So in the complete interval from 25 to 225, there are 50 integers divisible by 4; 24+3=27 of them are not divisible by either 6 or 10; 21+2=23 of them are divisible by 6 and/or 10.
So the probability that any multiple of 4 between 25 and 225 is not divisible by either 6 or 10 is 27/50.
Then the probability of choosing two numbers in that interval that are divisible by 4 but not divisible by either 6 or 10 is
Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! there are 201 numbers.
fifty of them (1/4) are divisible by 4
One third of those divisible by 4 are divisible by 6 as well, and two-thirds of 50 is 33
some additional are divisible by 10 and not 6 (40, 80, 100, 140,160,200, 220) leaving 26
the numbers left from 25-100 inclusive are
28/32/44/52/56/64/68/76/88/92 for 10 numbers
look now at
104/108/112/116/124/128/132/136/144/148/152/156/164/168/176/188/192/196, and 108, 132, 144, 156, 168, 192 are excluded, so there are 12 left
204/208/212/216/220/224 --exclude 204 and 216 for 4
The total number is 26
probability of choosing one is 26/201 and the second would be 25/200
that joint probability is 0.016
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