Nathan Lord to see the true capacity wow
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Front cheese
Definition: The probability of a random variable expectation
Several important expression
\ ((1 - x) \ sum ^ {n} _ {i = 0} x ^ i = 1 - x ^ {n + 1} \)
\(\sum^{inf}_{i = 0} x^i = \frac{1}{1 - x}\)
\(\sum^{n}_{i = 0} x^i = \frac{1 - x^{n + 1}}{1 - x}\)
\ (E (XY) = E (X) and (Y) \)
Desired linearity:
\ (E (the Y-the X-+) = E (the X-) + E (the Y-) \)
Note that this formula although the incident but no two events contribute to the desired requirements must be independent of
the next classical part of the problem will be mentioned
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skill
- First, prefixes and tips
For the discrete variables X, there are P (x == K) = P (x \ leq K) - P (x \ leq K - 1) = P (x \ geq K) - P (x \ geq K + 1)
With the latter may prove an important conclusion
to an occurrence probability P of the event,Repeated cross jumpUntil the number is expected to occur \ (\ frac {1} {
P} \) demonstrated the expression Similarly items before and after deployment at a second arc to equations (time to fill the hole)
- Second, the problem of the ball
For the problem is equivalent emotional understanding can equal probability
- Third, the classic problem
Directly attached to the courseware. .
Answer:
. 1, \ (\ ^ {n-SUM _} = {I}. 1 \ {n-FRAC {I}} \)
2、\(\frac{1}{i}\)
3、\(\sum^{n}_{i = 1}\sum^{n}_{j = 1, j != i} \frac{1}{ij} + \sum^{n}_{i = 1} \frac{1}{i}\)
4、\(\frac{1}{2}\)
5、(1) \(\frac{\tbinom{n}{m} (n - m)}{n!} = \frac{1}{m!}\)
(2)\(\frac{(n - m + 1)(n - m)!}{n!}\)
6、\(1 + \sum^{n}_{i = 2}\frac{a[i]}{a[1] + a[i]}\)
7、$$
8、\(\frac{2}{(j - i + 1)(j - i)}\)
9、
10、