An experiment is an occurrence whose result, or outcome is uncertain.

The set of all possible outcomes is called the sample space for the experiment.

Given a sample space S, an event E is a subset of S. The outcomes in E are called the favorable outcomes.

We say that E occurs in a particular experiment if the outcome of that experiment is one of the elements of E, that is, if the outcome of the experiment is favorable.

On-Line Tutorial Beginning With This Topic

1. Experiment: Cast a die and observe the number facing up.
    Outcomes: 1, 2, 3, 4, 5, 6
    Sample Space: S = {1, 2, 3, 4, 5, 6}
    An Event: E: the outcome is even; E = {2, 4, 6}

2. Here is an experiment that simulates tossing three fair distinguishable coins. To run the experiment, press the "Toss Coins" button and record the occurrences of heads and tails. The sample space is the set of eight possible outcomes:

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Let E be the event that heads comes up at least twice.

E = {HHH, HHT, HTH, THH}

If E and F are events in an experiment, then:

E' is the event that E does not occur.

E F is the event that either E occurs or F occurs (or both).

E F is the event that both E and F occur.

E and F are said to be disjoint or mutually exclusive if (E F) is empty.

Let S be the sample space for the coin tossing experiment in the box above, so that

   S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

Let E be the event that heads comes up at least twice;

   E = {HHH, HHT, HTH, THH},

   F = {HHT, HTH, HTT, THH, THT, TTH, TTT}.

Then:

   E' = {HTT, THT, TTH, TTT}
   E F = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
                = S
   E F = {HHT, HTH, THH}

E and F are not mutually exclusive, since E F ≠

If an experiment is performed N times, and the event E occurs fr(E) times, then the ratio

P(E)

=

fr(E) N

The number fr(E) is called the frequency of E. N, the number of times that the experiment is performed, is called the number of trials or the sample size.

If E consists of a single outcome s, then we refer to P(E) as the estimated probability of the outcome s, and write P(s).

On-Line Tutorial on Estimated Probability

In the above experiment (toss three coins) take E to be the event that heads comes up at least twice. To compute its estimated probability, let us use the simulated experiment below.

Every time you press "Toss Coins" the web page will compute both fr(E) and P(E).

Let S = {s₁, s₂, ... , s_n} be a sample space and let P(s_i) be the estimated probability of the event {s_i}. Then

(a) 0 ≤ P(s_i) ≤ 1
(b) P(s₁) + P(s₂) + ... + P(s_n) = 1
(c) If E = {e₁, e₂, ..., e_r}, then P(E) = P(e₁) + P(e₂) + ... + P(e_r).

In words:

(a) The estimated probability of each outcome is a number between 0 and 1.
(b) The estimated probabilities of all the outcomes add up to 1.
(c) The estimated probability of an event E is the sum of the estimated probabilities of the individual outcomes in E.

The theoretical probability, P(E), of an event E is its probability determined from the nature of the experiment rather than through actual experimentation.

The estimated probability approaches the theoretical probability as the number of trials gets larger and larger.

Notes

1. We write P(E) for both estimated and theoretical probability. Which one we are referring to should always be clear from the context.
2. Theoretical probability satisfies the same properties (shown above) as estimated probability.

In the above experiment, there are eight outcomes in S, and half of them are in E. Therefore, we expect E to occur half the time. In other words, the theoretical probability of E is 0.5.

If you "toss the coins" in the simulated experiment arriba a large number of times, the estimated probability should approach the theoretical probabilty of 0.5. In the following simulation, you can toss the coins 50 times with each click on the button.

In an experiment in which all outcomes are equally likely, the probability of an event E is given by

P(E)	=	number of favorable outcomes total number of outcomes
	=	n(E) n(S)

In the above experiment (toss three coins) there are eight equally likely outcomes in S, and half of them are in E (the event that heads comes up at least twice). Therefore,

P(E)	=	n(E) n(S)
P(E)	=	4 8 = 1 2 .

Since the different kinds of probability share similar properties, we refer to estimated and theoretical probability collectively as just probability. What they have in common is the idea of a probability distribution:

A finite sample space is just a finite set S. A probability distribution is an assignment of a number P(s_i) to each outcome s_i in a sample space S ={s₁, s₂, ... , s_n} so that

  (a) 0 ≤ P(s_i) ≤ 1
  (b) P(s₁) + P(s₂) + ... + P(s_n) = 1.

P(s_i) is called the probability of s_i. Given a probability distribution, we obtain the probability of an event E by adding up the probabilities of the outcomes in E.

If P(E) = 0, we call E an impossible event. The event is always impossible, since something must happen.

Notes

1. All the above properties apply to both estimated and theoretical probability. Thus, when we speak only of "probability," we could mean either estimated or theoretical probability, depending on the context.

On-Line Tutorial on Probability Distributions

1. All the examples of estimated and theoretical probability above give examples of probability distributions.

2. Let us take S = {H, T} and make the assignments P(H) = 0.2, P(T) = 0.8. Since these numbers are between 0 and 1, and add to 1, they specify a probability distribution.

3. With S = (H, T} again, we could also take P(H) = 1, P(T) = 0, so that T is an impossible event.

4. The following table gives a probability distribution for the sample space S = {1, 2, 3, 4, 5, 6}.

It follows that

P({1, 6}) = 0.3 + 0.1 = 0.4

P({2, 3}) = 0.3 + 0 = 0.3

P(3) = 0         An impossible event

5. Suppose we toss three coins as above, but this time, we only look at the number of heads that come up. In other words, S = {0, 1, 2, 3}. The (theoretical) probability distribution is given by counting the number of combinations that give 0, 1, 2, and 3 heads:

The following simulation computes the estimated probability distribution of the above experiment. You will find that, after many coin tosses, these probabilities converge to the theoretical ones shown above.

Mutually Exclusive Events
If E and F are mutually exclusive events, then

P(E F) = P(E) + P(F).

P(E) = P(E₁) + P(E₂) + . . . + P(E_n).

General Addition Principle
If E and F are any two events, then

P(E F) = P(E) + P(F) - P(E F).

Let S be the sample space for the coin-tossing experiment above;

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Let E be the event that heads comes up at exactly once;

E = {HTT, THT, TTH}

F = {HHT, HTH, THH}.

Then E and F are mutually exclusive, and

P(E F) = P(E) + P(F) = 3/8 + 3/8 = 6/8.

Now let E be as above, and let F be the event that heads comes up at most once:

F = {HTT, THT, TTH, TTT}

P(E F) = P(E) + P(F) - P(E F)
      = 3/8 + 4/8 3/8 = 4/8.

The following are true for any sample space S and any event E.

P(S) = 1	The probability of something happening is 1.
P() = 0	The probability of nothing happening is 0.
P(E') = 1-P(E)	The probability of E not happening is 1 minus the probability of E.

Continuing with

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Let E be the event that heads comes up at exactly once;

E = {HTT, THT, TTH}.

P(E') = 1- P(E) = 1- 3/8 = 5/8.

If E and F are two events, then the conditional probability, P(E | F), is the probability that E occurs, given that F occurs, and is defined by

P(E | F)

=

P(E F) P(F)

.

We can rewrite this formula in a form known as the multiplication principle:

P(E F) = P(F)P(E | F).

Conditional Estimated Probability
If E and F are events and P is estimated probability, then

P(E | F)

=

fr(E F) fr(F)

.

Conditional Probability for Equally Likely Outcomes
If all the outcomes in S are equally likely, then

P(E | F)

=

n(E F) n(F)

.

On-Line Tutorial Beginning With This Topic

Let S be the original sample space for the experiment above;

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Let E be the event that heads comes up at exactly once;

E = {HTT, THT, TTH}

F = {HHH, HHT, HTH, HTT}.

Then the probability that heads comes up exactly once, given that the first comes up heads is

P(E | F)	=	P(E F) P(F)
	=	P{HTT} P(F)
	=	1/8 1/2 = 1 4 .

Since the outcomes in this experiment are equally likely, we could also use the formula

P(E | F)	=	n(E F) n(F)
	=	1 4 .

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The events E and F are independent if

P(E | F) = P(E)

If two events E and F are not independent, then they are dependent.

Given any number of mutually independent events (that is, each one of them is independent of the intersection of any combination of the others), the probability of their intersection is the product of the probabilities of the individual events.

As in the example immediately above, let S be the original sample space for the experiment above;

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Let E be the event that heads comes up at exactly once;

E = {HTT, THT, TTH}

F = {HHH, HHT, HTH, HTT}.

To test these two events for independence, we check the formula

P(E F) = P(E)P(F).

P(E) = 3/8, P(F) = 1/2, and
E F = {HTT}, so that P(E F) = 1/8.

(3/8)(1/2) 1/8,

The short form of Bayes' Theorem states that if E and F are events, then

P(F | E)

=

P(E | F)P(F) P(E | F)P(F) + P(E | F')P(F')

.

We can often compute P(F | E) by instead constructing a probability tree. (To see how, go to the tutorial by following the link below.)

An expanded form of Bayes' Theorem states that if E is an event, and if F₁, F₂, and F₃ are a partition of the sample space S, then

P(F1 | E)

=

P(E | F1)P(F1) P(E | F1)P(F1) + P(E | F2)P(F2) + P(E | F3)P(F3)

.

A similar formula works for a partition of S into four or more events.

On-Line Tutorial Beginning With This Topic

If P(E | F) = 0.95     P(E | F') = 0.15     P(F) = 0.1     P(F') = 0.9,   then

P(F | E)	=	P(E | F)P(F) P(E | F)P(F) + P(E | F')P(F')
	=	(0.95)(0.1) (0.95)(0.1) + (0.15)(0.9)
		0.4130.

This example comes from a scenario discussed in the tutorial (link on the adjacent box).

A Markov system (or Markov process or Markov chain) is a system that can be in one of several (numbered) states, and can pass from one state to another each time step according to fixed probabilities.

If a Markov system is in state i, there is a fixed probability, p_ij, of it going into state j the next time step, and p_ij is called a transition probability.

A Markov system can be illustrated by means of a state transition diagram, which is a diagram showing all the states and transition probabilities. (See example opposite.)

The matrix P whose ijth entry is p_ij is called the transition matrix associated with the system. The entries in each row add up to 1. Thus, for instance, a 22 transition matrix P would be set up as in the following figure.

Transition Diagram:
(Missing arrows indicate zero probability.)

Matrix:

A distribution vector is a row vector with one nonnegative entry for each state in the system. The entries can represent the number of individuals in each state of the system.

A probability vector is a row vector in which the entries are nonnegative and add up to 1. The entries in a probability vector can represent the probabilities of finding a system in each of the states.

If v is an initial distribution vector and P is the transition matrix for a Markov system, then the distribution vector after 1 step is the matrix product, vP.

Distribution After 1 Step:   vP

Distribution After 2 Steps:   vP²

Distribution After n Steps:   vPⁿ

P² is the two-step transition matrix for the system. Similarly, P³ is the three-step transition matrix, and Pⁿ is the n-step transition matrix. This means that the ijth entry in Pⁿ is the probability that the system will pass from state i to state j in n steps.

Try our on-line utility which computes powers of the transition matrix and its action on the distribution vector. Even better (and far more flexible) is our Matrix Algebra Tool, (updated in November 2003) where you can compute several things simultaneously, have the answers shown in fraction or decimal form, and also compute inverses and even complicated expressions involving several matrices.

Let

P =		0.2	0.8	0
		0.4	0	0.6
		0.5	0.5	0

and let v = [ 100   200   300 ] be an initial distribution. Then the distribution after one step is given by

vP = [ 100   200   300 ]		0.2	0.8	0
		0.4	0	0.6
		0.5	0.5	0
= [ 250   230   120 ]

The distribution one step later is given by

vP2 = (vP)P
= [ 250   230   120 ]		0.2	0.8	0
		0.4	0	0.6
		0.5	0.5	0
= [ 202   260   138 ].

To obtain the two-step transition matrix, we calculate

P2 =		0.2	0.8	0			0.2	0.8	0
		0.4	0	0.6			0.4	0	0.6
		0.5	0.5	0			0.5	0.5	0

=		0.36	0.16	0.48
		0.38	0.62	0
		0.3	0.4	0.3

Thus, for example, the probability of going from State 3 to State 1 in two steps is given by the 3,1-entry in P², namely 0.3.

If P is a transition matrix for a Markov system, and if v is a distribution vector with the property that vP = v, then we refer to v as a steady state (distribution) vector.

To find a steady state distribution for a Markov System with transition matrix P, we solve the system of equations given by

x + y + z + . . .	=	1
[x   y   z . . . ]P	=	[x   y   z . . .]

v = [x   y   z . . . ]

Long Term Behavior If the higher and higher powers of P approach a fixed matrix P, we refer to P as the steady state or long-term transition matrix. If a Markov system is regular, then its long-term transition matrix is given by the square matrix whose rows are all the same and equal to the steady state probability vector

[x   y   z . . .].

Using technology, it is often possible to approximate P with great accuracy by simply computing a very large power of P. How large? You know it is large enough when the rows are all the same to withing the accuracy you desire. For small matrices like the ones in the textbook, P²⁵⁶ is usually more than good enough. (256 is a power of 2, and so the computation of P²⁵⁶ is especially fast on our Matrix Algebra Tool. Try it!) Warning;If the matrix does not stabalize fairly quickly, then there is a danger of significant computational: the larger the numer of computations required, the more serious this problem becomes.

The transition matrix

P =		0.2	0.8	0
		0.4	0	0.6
		0.5	0.5	0

that we looked at above is regular, since P³ contains only non-zero entries. (Check this on your graphing calculator or on our on-line utility.)

The steady state probability vector is given by

v = [35/99   40/99   24/99],

vP = [35/99   40/99   24/99]		0.2	0.8	0
		0.4	0	0.6
		0.5	0.5	0
= [35/99   40/99   24/99]
= v.

The long-term transition matrix is therefore

P =		35/99	40/99	24/99
		35/99	40/99	24/99
		35/99	40/99	24/99

(To check this, calculate P²⁵⁶ on the on-line utility.)

Using the Matrix Algebra Tool

Open the Matrix Algebra Tool and enter the matrix P using the following format (note the commas between each pair of entries in each row.): (If you like, you can just copy the blue text below and paste it into the window.)

P = [0.2, 0.8, 0
0.4, 0, 0.6
0.5, 0.5, 0]