Problems on first four moments of binomial distribution
Answers
NUMBER OF POSSIBLE SEQUENCES FOR N TRIALS. If any one of K
mutually exclusive and exhaustive events can occur on each of N trials, then there are Kn
different sequences that may result from a set of such trials.
2. SEQUENCES. If K1,...,KN are the numbers of distinct events that can occur
on trials 1,..., N in a series, then the number of different sequences of N events that can
occur is (K1)(K2)...(KN).
3. PERMUTATIONS The number of different ways that N distinct things may
be arranged in order is
N! = (1)(2)(3)...(N-1)(N), (where 0! = 1).
3B. PERMUTATIONS OF SIMILAR OBJECTS. Suppose we have N objects, N1
alike, N2 alike,..., Nk alike (ΣNi = N). Then, the number of ways of arranging these objects
is
N ! N !...N !
N!
1 2 k
4. ORDERED COMBINATIONS; or, PERMUTATIONS OF N OBJECTS TAKEN r
AT A TIME. The number of ways of selecting & arranging r objects from among N distinct
objects is
(N -r)!
N!
5. COMBINATIONS. The total number of ways of selecting r distinct
combinations of N objects, irrespective of order, is
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N -r
N = r
N = r!(N -r)!
N!
6. Binomial distribution. In sampling from a stationary Bernoulli process, with
the probability of success equal to p, the probability of observing exactly r successes in N
independent trials is
p q r!(N -r)!
N!
p q = r
N r N-r r N-r
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7. Binomially distributed variables. Let Xi = 1 if the ith bernoulli trial is
successful, 0 otherwise. If X = ΣXi, where the Xi’s are independent and identically
distributed (iid), then X has a binomial distribution, and E(X) = Np, V(X) = Npq.
The Binomial Distribution