the product of n consecutive positive integers from n to 1 is n!
n! = n(n-1)(n-2)(n-3)…3.2.1
note: 0! is defined as 1.
Permutation:
each of the ordered subsets which can be formed by selecting some or all of the elements of a set;
the multiplication principle:
if one operation can be performed in m different ways, and when it has been performed in any of these ways, a second operation can then be performed in n different ways, the number of ways of performing the two operations is m x n
eg. if there are 3 different main courses & 4 different desserts, you have a choice of 3×4=12 different two course meals
nPr:
the no. of arrangements of n different objects taken r at a time;
= n!/(n-r)!;
Arrangements in a circle:
the no. of ways of arranging n different objects in a circle, regarding clockwise & anticlockwise as different:
= (n-1)!;
Arrangements of n objects in a row, when not all are different:
if p alike of one kind, q alike of another kind, etc..
= n!/(p!q!…);
eg. how many ways can the letters of the word mammal be rearranged to make different words?
n = 6; p = 2 for letter a; q = 1 for letter l; r = 3 for letter m;
thus = 6! / (2! x 1! x 3!)
Arrangements with restrictions:
restriction principle: always fill a restriction first
eg. number of ways of arranging in a row 6 men and 2 boys:
if 2 boys must be together:
regard the boys as 1 unit, thus 7 objects not 8
thus 7! ways, but as the 2 boys can be arranged 2! ways amongst themselves ⇒ 2!7!
if 2 boys must NOT be together:
number arrangements without restriction = 8!
⇒ answer = 8! - number arrangements 2 boys are together ⇒ 8! - (2!7!)
if there must be at least 3 men separating the boys:
calculate the number of ways of arranging the boys alone:
sum up number of arrangements for boy B when boy A is placed in each of the 8 positions = 20
use multiplication principle to determine how the remaining 6 men can be arranged = 20 x 6!
Combination:
each of the subsets which can be formed by selecting some or all of the elements of the set without regard to the order in which the elements appear in the subset;
nCr:
the no. of combinations of n different objects taken r at a time;
= nPr/r! = n!/[r!(n-r)!];
Mutually exclusive operations:
when the selection of one object eliminates the possibility of it being selected again in that arrangement;
if two operations are mutually exclusive then the no.of arrangements possible with each are added (not multiplied) to obtain the total no. of possible arrangements;
ie. intersection A & B is a null set,
if two or more events cannot occur same time, Pr(A or B) = Pr(A) + Pr(B); (addition principle)
Basic probability definitions
Event: a set of favourable outcomes;
Trial: eg. the tossing of a die;
Sample space: E = all possible outcomes;
Probability of outcomes corresponding to A events:
assume all sample points equally likely,
Pr(A) = [no. outcomes A]/[total no. possible outcomes];
Pr(A or B) = Pr(A) + (Pr(B) - Pr(A&B);
thus, the probability of drawing an Ace or a Heart from a pack of cards = 1/13 + 1/4 - 1/52 = 16/52 = 4/13
Independent events:
A & B are independent if:
Pr(A&B) = Pr(A).Pr(B);
Conditional Probability:
Pr(B) given A = Pr(B/A) = Pr(A&B)/Pr(A);
if A,B are independent, then:
Pr(B/A) = [Pr(A).Pr(B)/Pr(A)] = Pr(B);
Baye's theorem
a method of UPDATING you pre-existing probability of an event based upon new data
same as for binomial except there is sampling without replacement;
variables:
N = size of pop.;
n = size of sample;
D = no. of kind A in pop.;
N-D = no of kind B in pop.;
X = no. of kind A in sample;
Pr(X=x) = nCr(D#x).nCr(N-D#n-x)/nCr(N#n)
= Pr(sample contains x of kind A),
u = nD/N
s'^2 = nD(1-D/N)(N-n)/[N(N-1)],
if N is very large, & n small, can approx. to binomial distribution;
Poisson distribution:
when to use:
as an approx. to binomial distrib. when p is very small and n is very large;
when the no. of times an event occurs can be counted but there is no upper limit to the no. of times it may occur;
Equation:
Pr(X=x) = e^(-u) * u^x / x!,
Eg. On average there are 2.5 cars per quarter-hour at a petrol station, what is the prob. that during a particular quarter-hour there will be some cars at the petrol station?
u = 2.5, Pr(X>=1) = 1 - Pr(X=0),
Pr(X=0) = e^(-2.5) ⇒ Pr(X>=1) = 0.92;
Normal Approx.:
u = u, s' = SQR(u),
Good approx. if u is large;
Eg. On average, there are 20 people asking for an item each week, what is the minimum no. of items the store must have in stock each week to be almost certain of not having to refuse demand for this item?
u = 20, s' = SQR(20),
Min. no. items = u + 3s' = 34;
Hypothesis testing:
Null hypothesis(H0): that there is no effect of one variable on another;
Alternative hypothesis(H1): that there is an effect;
H1 is likely to be true if the results are very unlikely to have been obtained if H0 were true;
Significance level:
p = Pr(obtaining obs. as extreme as the ones obtained if H0 were true),
Type I error:
a = Pr(deciding to reject H0 when H0 true);
Type II error:
b = Pr(accepting H0 when H1 is true);
if p ⇐ a, then reject H0,
if p > a, then accept H0,
Usually a is designated 0.05;
Z-test:
If H0 is a standard distribution:
p =Pr(z ⇐ (x-u0)(SQRn)/s') or Pr(z >= (x-u0)(SQRn)/s')
(x < u0) (x > u0)
where, u0,s' are mean, sd of H0 true, n = size of sample, x = mean of sample;
the pop. should be approx. of normal distrib., if it is skewed, a large sample is necessary for the z-test to be true;