Need not for speed, but for Prime
An integer is randomly selected in the interval between 1 and 1 0 1 0 inclusive. What is the probability that it is a prime number ?
Give your answer to 2 significant figures.
The answer is 0.0455052511.
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1 solution
ln ( n ) n is just an approximation and is off the actual value. There are better approximations, which can give you a different value.
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I did not know that. Could you site some examples?
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x / ln x gives an approximation because x → ∞ lim x / ln x π ( x ) = 1 . However, for any finite value, there's always some percentage of error, which decreases as x grows. OEIS has a nice table on this. Check this page which discusses a lot of other prime counting functions with much smaller margins of error.
ln 1 0 1 0 1 0 1 0 ≈ 4 3 4 2 9 4 4 8 2
while the actual value is 455,052,511. Which means that this approximation is off by 20758029.
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@Arulx Z – Thanks for enlightening. :-)
I've edited the problem starement. Please check if its befitting now.
I am changing this to a Number Theory problem. Is that alright?
@Agnisom Chattopadhay: Its fine. Thanks.
According to Gauss' Prime Counting Function , number of primes less than n, π ( n ) = l n ( n ) n And its obvious that there are n numbers in this range.
So, the probability, p = n π ( n ) = n l n ( n ) n = l n ( n ) 1 For our problem, n = 1 0 1 0
Now its just the calculator's job. Thus, p = l n ( 1 0 1 0 ) 1 = 0 . 0 4 3 4 3 ( a p p r o x . )