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Old 09-24-2014, 09:42 AM   #9881
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Originally Posted by sickman411
Ah, right. You made a small mistake when you applied the logarithms, and I hadn't noticed. Don't forget that only x is raised to the power x, not 2. The expression you have there is for y = (2x)^x, not y = 2 x^x

Here, I couldn't be arsed to type it all out and wouldn't look as neat anyway.


Yeah, this is spot on, no point in me posting the same thing.
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Old Yesterday, 12:33 AM   #9882
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Probability theory question:

Suppose that three numbers are selected one by one, at random, and without replacement, from the set of integers {1, 2, 3, . . . , n}. What is the probability that the third number falls between the first two if the first number is smaller than the second?
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Old Yesterday, 01:46 AM   #9883
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There are n(n-1)(n-2) different ways to select three numbers out of that set.

Suppose n_1 is the Nth number and n_2 is the Mth number, then there are only M-N-1 number choices left for n_3.
Sum over all M, N for M > N > 0, M < or equal to n.

Divide by n(n-1)(n-2).

edit: actually that completely ignores that the first number is smaller than the second one. So, what you want is P(*that event* | n_1 < n_2) and that is P(*that event* AND n_1 < n_2)*P(n_1 < n_2)

P(n_3 falls between n_1 and n_2) is what I found, now you need P(n_1 < n_2).
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