Probability and Recipricals ```Name: Safwaan K. Status: student Age: N/A Location: N/A Country: N/A Date: N/A ``` Question: Hi, I have made a bit of a discovery that i wish to share with you. Could you let me know if this is something new that I have discovered. If it is, I want all the credit :) Law Of Probable Recipricol Division: Replies: Bascially, I have found that the probability of selecting any random number which is exactly divisible by itself is exactly equal to the recipricol of that number. That is to say: The probability that X is exactly divisble by Y is equal to 1/x. or the probability that (X MOD Y = 0) = 1/x for example: What is the probability that a random number between 1 and 100000000 is divisible by 19? The answer is 1/19 = 0.05263, i.e there is a 5.263% chance that a random number is divisible by 19. Probability is not really my field, and I have not been able to dis-prove this, but I would simply qualify the original stated example by adding, rather than "between", the words "numbers from 1 to 10,000,000 inclusive". The solution could fall apart if the end values are not included (I suspect). I note that the white filling of an Oreo cookie lies 'between' the two chocolate wafers but does not include them. If your solution is intended to include the end digits, it is better stated; otherwise, the cookie crumbles. ;) Thanks for using NEWTON! Ric Rupnik I am not sure I find your result a surprise. In the interval between 1 and 100,000,000 which =100x10^6 there 100x10^6 / 19 = 5.2361578... x10^6 multiples of 19. 1x19, 2x19, 3x19, ..., 5.23631578...x10^6x 19 = 100x10^6 give-or-take round off error. So the probability of randomly choosing a number in the interval between the integers 1 and 100x10^6 that is a multiple of 19 is: the number of multiples of 19 in the interval divided by the number of integers in the interval, or 5.2361578...x10^6 / 100x10^6 = 0.052361578... Vince Calder Click here to return to the Mathematics Archives

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