Let S = {o_{1},o_{2},o_{3},o_{4},o_{5}} Let E_{1} = {o_{1},o_{2},o_{3}} Then E_{1}' = {o_{4},o_{5}}
Let E_{1} = {o_{1},o_{3},o_{5}} Let E_{2} = {o_{1},o_{2},o_{3}} Then E_{1} ∩ E_{2} = {o_{1},o_{3}}
Let E_{1} = {o_{1},o_{3},o_{5}} Let E_{2} = {o_{1},o_{2},o_{3}} Then E_{1} ∪ E_{2} = {o_{1},o_{2},o_{3},o_{5}}
Let E_{1} = {o_{1},o_{2},o_{3}} Let E_{2} = {o_{4},o_{5}} Then E_{1} ∩ E_{2} = {} = ∅
Let E_{1} = {o_{1},o_{2}} Let E_{2} = {o_{4},o_{5}} Let E_{3} = {o_{3}} Then E_{1} and E_{2} and E_{3} are pairwise disjoint.
Let S = {o_{1},o_{2},o_{3},o_{4},o_{5}} Let E_{1} = {o_{1},o_{2}} Let E_{2} = {o_{4},o_{5}} Let E_{3} = {o_{3}} Then E_{1} and E_{2} and E_{3} form a partition of sample space S.
Properties of a Partition of the Sample Space:Categorical  Numerical 

Values that cannot be measured using measuring tools
The color, shape, or other discrete property of a sampled item A yesno, multiple choice, or integer outcome of a random event (e.g. a rolled die, playing card, genetic trait, or drawn ball) 
Values that can be measured using instruments
Length, volume, mass, or other measured value of sampled items Any other value that can be measured with instruments (e.g. temperature, density, pressure, or other physical property) 
Count  Portion 
Number of occurrences of values that cannot be measured
How many instances have the color, shape, or other property A nonnegative integer number of outcomes of a random event (e.g. number of heads, number with the trait, or ball count) 
The portion of a given value as a fraction of that value
Probabilities are portions of certainty  Fractions of wholes Restricted to these values: 0 ≤ p ≤ 1 (e.g. probability, 1/2 full, or part of any measured value) 
Replacement  No Replacement 

Objects or outcomes are put back before the next drawing.
An infinite supply of objects is available. Outcomes are part of a permanent random device. (e.g. a flipped coin, a rolled die, or a spinner) 
Objects or outcomes become unavailable or disallowed once used.
Race and tournament finishing orders Objects are kept or discarded. (e.g. a dealt hand of cards, books on a shelf, or a committee) 
Order Matters  No Order 

Outcomes are noted or arranged as they are picked.
Race and tournament finishing orders. Outcomes are used for different purposes. (e.g. president, vice president, secretary) 
Objects or outcomes are chosen simultaneously.
Rearranging objects does not change the outcome. Objects are dumped into a holder. (e.g. a hand of cards, colored checkers, or a committee) 
Note that counts are not numerical data.
The following are methods of counting the numbers of items and outcomes in everyday life:
Let E_{1} = {o_{1},o_{3},o_{5}} Then n(E_{1}) = 3} Note: N(∅) = 0
Let S = {o_{1},o_{2},o_{3},o_{4},o_{5}}
Let E_{1} = {o_{1},o_{3},o_{5}}
Let E_{2} = {o_{1},o_{2},o_{3}}
So E_{1} ∩ E_{2} = {o_{1},o_{3}}
Then n(E_{1} ∪ E_{2}) = n(E_{1}) + n(E_{2}) −
n(E_{1} ∩ E_{2}) = 3 + 3 − 2 = 4
Let S = {o_{1},o_{2},o_{3},o_{4},o_{5}}
Let E_{1} = {o_{1},o_{2},o_{3}}
Let E_{2} = {o_{4},o_{5}}
So E_{1} ∩ E_{2} = ∅
Then n(E_{1} ∪ E_{2}) = n(E_{1}) + n(E_{2})
= 3 + 2 = 5
Let S = {o_{1},o_{2},o_{3},o_{4},o_{5}} Let E_{1} = {o_{1},o_{2},o_{3}} Then n(E_{1}') = n(S) − n(E_{1}) = 5 − 3 = 2
Let n(S_{1}) = 4 Let n(S_{2}) = 3 Let n(S_{3}) = 5 Then n(S) = 4 • 3 • 5 = 60
Let n(S_{1}) = n(S_{2}) = n(S_{3}) = 5 Then n(S) = 5^{3} = 125
This works when drawn objects are replaced, and the order of the outcomes matters.Where there are n objects to choose from, and r choices are made with replacement, and order matters, the number of outcomes is n^{r}
Let n = 5. Then n! = 5!  5 • 4 • 3 • 2 • 1 = 120
This works when drawn objects are not replaced, the order of the outcomes matters, and all of the objects are taken.Then A(n,r_{1},r_{2},...,r_{g}) =  n! 
r_{1}! r_{2}! ... r_{g}! 
Example: Let n = 5 Let g = 3 Let r_{1} = 2 Let r_{2} = 1 Let r_{3} = 2
Then A(n,r_{1},r_{2},r_{3}) =  n!  =  5!  = 30 
r_{1}! r_{2}! r_{3}!  2! 1! 2! 
This works when drawn objects are not replaced, the order of the objects matters, some objects are indistinguishable from others, and all of the objects are taken.
Let n = 5 Let r = 3
Then P(n,r) =  n!  =  5!  = 60 
(n−r)!  (5−3)! 
This works when drawn objects are not replaced, and the order of the outcomes matters.
Let n = 5 Let r = 3
Then C(n,r) =  n!  =  5!  = 10 
r! (n−r)!  3! (5−3)! 
This works when drawn objects are not replaced, and the order of the outcomes does not matter.
Let n = 5 Let r = 3
Then R(n,r) =  (n+r−1)!  = C(n+r−1,r) =  5!  = 35 
r! (n−1)!  3! (51)! 
This works when drawn objects are replaced, and the order of the outcomes does not matter.
NOTE: This is rare in probability. It is included for studying unusual counting problems that do occur in daily life. Special methods (discussed later) are needed to obtain equally likely outcomes in this kind of experiment.OBJECTS REPLACED 
OBJECTS NOT REPLACED 


ORDER MATTERS 
Power Principle n^{r} 
Permutation P(n,r) 
NO ORDER 
Recombination R(n,r) 
Combination C(n,r) 
The following are probability measures that occur in everyday life.
It is not nearly as useful as probability.
Let E_{i} = {o_{j},o_{j+1}, ... ,o_{k}}
Then Pr(E_{i}) = w_{j} + w_{j+1} + ... + w_{k}w_{i} = 1 / n(S)
Pr(E_{i}) = n(E_{i}) / n(S)Note that, since E ∩ F occurs twice, once in E, and also once in F, one of them must be subtracted out.
Note that, since E ∩ F is empty in disjoint events, its probability is 0, and it does not have to be subtracted out.
The probability of the event A, given that B has occurred, is the probability that both events occurred, divided by the probability that event B occurred.
Pr(S_{1}A) =  Pr(AS_{1}) Pr(S_{1}) 
Pr(AS_{1}) Pr(S_{1}) + Pr(AS_{2}) Pr(S_{2}) + ... + Pr(AS_{n}) Pr(S_{n}) 
Note that the following distributions use categorical data:
Bernoulli Process  A Bernoulli process is a repetition of the same operation a specified number of times.
It has the following properties:
Explanation of the various parts:

Example:
Let n = 5
Let r = 3
Let p = 1/4
Then Pr(3 successes) = C(5,3) 1/4^{3} (3/4)^{2} = 10 (1/64) (9/16) = 90 / 1024 = 45 / 512
The Partitioning Process  The partitioning process is used to find probabilities where several kinds of
objects can be selected, but where replacement does not occur.
It has the following properties:
Explanation of the various parts:

Pr(r successes) =  C(k,r) C(jk,nr) 
C(j,n) 
Example:
Let j = 5 in the pool
Let k = 3 are successes
Let n = 4 drawn
Let r = 2 successes drawn
Then Pr(2 successes) =  C(3,2) C(2,2)  =  3 • 1  =  3 
C(5,4)  5  5 
The Partitioning Process can also be used with multiple kinds of identical objects, but it is no longer
a distribution.
It has the following properties:
Explanation of the various parts:

Pr(o_{r1,r2,...,rg}) =  C(k_{1},r_{1})•C(k_{2},r_{2})• ... •C(k_{g},r_{g}) 
C(j,n) 
Example:
Let j = 5 in the pool
Let g = 3 kinds
Let k_{1} = 2 of kind 1
Let k_{2} = 1 of kind 2
Let k_{3} = 2 of kind 3
Let n = 4 drawn
Let r_{1} = 2 of kind 1 drawn
Let r_{2} = 1 of kind 2 drawn
Let r_{3} = 1 of kind 3 drawn
Pr(2,1,1) =  C(2,2)•C(1,1)•C(2,1)  =  1 • 1 • 2  =  2 
C(5,4)  5  5 
Note that, although real numbers are used, the outcomes are still categorical data because only certain values can occur.
The binomial random variable is used where replacement occurs.
The hypergeometric random variable is used where replacement does not occur.
Some books use this, but do not identify it by name.
Example: A dice game at a carnival costs $3 per play. If you win, you get $4, if you lose, you get nothing. A win is a
roll of 2, 3, 4, or 5.
Here is a chart of the random variable of this game:
Die roll  Value  Weight 

1  $3  1/6 
2  +$1  1/6 
3  +$1  1/6 
4  +$1  1/6 
5  +$1  1/6 
6  $3  1/6 
Example, taken from the random variable example above:
Die roll  Value  Weight  Product 

1  $3  1/6  $3/6 
2  +$1  1/6  +$1/6 
3  +$1  1/6  +$1/6 
4  +$1  1/6  +$1/6 
5  +$1  1/6  +$1/6 
6  $3  1/6  $3/6 
Total  1  $2/6 
Die roll  Value  Weight  Product  Deviation  Squared Dev  Product 

1  $3  1/6  $3/6  16/6  64/9  32/27 
2  +$1  1/6  +$1/6  +8/6  16/9  8/27 
3  +$1  1/6  +$1/6  +8/6  16/9  8/27 
4  +$1  1/6  +$1/6  +8/6  16/9  8/27 
5  +$1  1/6  +$1/6  +8/6  16/9  8/27 
6  $3  1/6  $3/6  16/6  64/9  32/27 
Total  1  $2/6  96/27 
Note that the fractions obtained in the process were reduced, but just enough to obtain a common denominator.
In the example above, the standard deviation is √(32/9) = 4/3 √2 , or approximately $1.89.
Multiply the number of trials by the probability of success.
Var(x) = n p (1p)
σ = √(n p (1p))
E(x) = k n / j
Var{x} = n k (ik) / j^{2}
σ = √(n k (ik) / j^{2})
Note that the above is simply applying the normalizing equation Z = (Xμ)/σ by substituting the variables j and k for X.
A sample case: How many colors can a light panel show by changing lightbulbs?
R(3,6) =  (n+r−1)!  =  (3+6−1)!  = 28 
r! (n−1)!  6! (31)! 
The actual 28 outcomes are:
color  Outcomes  
1  RRRRRR red  GGGGGG green  BBBBBB blue  
2  RRRRRG vermillion  RRRRGG amber  RRRGGG yellow  RRGGGG chartreuse  RGGGGG leaf  
RRRRRB carmine  RRRRBB cerise  RRRBBB magenta  RRBBBB purple  RBBBBB violet  
GGGGGB kelly green  GGGGBB aquamarine  GGGBBB cyan  GGBBBB sky blue  GBBBBB azure blue  
3  RGBBBB light blue  RGGBBB light azure  RGGGBB light aqua  RGGGGB light green  RRGBBB lilac  
RRGGBB white  RRGGGB light chart  RRRGBB rose  RRRGGB peach  RRRRGB pink 
There are 3 cases of one color, 15 cases of two colors, and 10 cases of three colors. Each case is a unique color mixture.
There are 3 cases of one color, 186 cases of two colors, and 540 cases of three colors.
The extra cases are unwanted duplicate colors with the bulbs switched around in the sockets.
The correct numbers of outcomes are there.
Because different bins contain different numbers of bulbs, tree branches have different probabilities.
The setup has to be changed for each case.
The procedure is:
Some advantages:
Some disadvantages:
The procedure
The number of gaps is always one less than the number of choice balls drawn.
Note: a gap can appear at either one or both ends of the choice levels.
The balls in the tray now contain the desired choice. All outcomes are equally likely.
Using a ball set R G B 1 2 3 4 5 with "$" as a gap.
Examples:
Drawing  Gap Finding  Move Choices  Final Outcome 

G12345  → G12345  → G12345  → GGGGGG  green 
GB1245  → GB12$45  → G12B45  → GGGBBB  cyan 
RG1234  → RG1234$  → R1234G  → RRRRRG  vermillion 
RG2345  → RG$2345  → RG2345  → RGGGGG  leaf 
RGB123  → RGB123$$  → R123GB  → RRRRGB  pink 
RGB135  → RGB1$3$5  → R1G3B5  → RRGGBB  white 
RGB245  → RGB$2$45  → RG2B45  → RGGBBB  light azure 
RGB345  → RGB$$345  → RGB345  → RGBBBB  light blue 
RGB145  → RGB1$$45  → R1GB45  → RRGBBB  lilac 
RGB124  → RGB12$4$  → R12G4B  → RRRGGB  peach 
Some advantages:
Some disadvantages:
Links: