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a background to understanding probability is understanding set theory, as venn
diagrams and intersections/unions are often used to denote the events we need
to figure to determine probability.
"The probability of an event A is the sum of the weights of all sample
points in A. Therefore 0 <= P(A) <= 1 and P(null-set) = 0 and P(S) = 1".
Meaning, it is the proportion of times that that event will occur
"furthermore, if A1 A2 A3 is a sequence of mutually exclusive events then P(A1
union A2 union A3 union ... ) = P(A1) + P(A2) + P(A3) ..... " since the sum of
all the sample points is 1 the addition of the probabilities which give the
event in question will yield the total probability of that event. Another way
of expressing this is to say that it is the number of times that divided by
the total events in the sample space. It is the part over the whole.
Probability is "the probablity of an event E is defined as E/S where S is the sample space"
probability is always a number between 0 and 1 and is usually a fraction of
part/whole. Do not confuse with counting the sample space (see counting.txt).
The counting the sample space gives you the "whole" part of this equation.
The "part" is the sample space which fulfills the condition you are
investigating.
1aconditional probability: Let E1 and E2 be any 2 events of an experiment. The conditional probability of E1 given E2, denoted by P(E1|E2) is defined as P(E1 INTERSECTION E2)/P(E2)
a coin is tossed 6 times. find the probability that 4 heads will be gotten.
The probability of getting 4 heads and 2 tails is (1/2)2 = 1/64
The total number of choices where there is a head are C(n,r) = C(6,4) 6!/4!2! = 6*5/2 = 15
therefore the probability is 15/64
part over the whole
pdf is probability density function which is the set of all real numbers (x,
the random variable) for the continuous random variable X
pmf is probability mass function (also known as simply the probability
function and probability distribution) of a
random variable X as a function of an instance of that random variable x
cumulative distribution: denoted by F(x), this is defined by
F(x) = P(X <= x) = sum from t<= x of f(t)
There are cumulative distributions of both continuous random variables and
discrete random variables.
examples of probability distributions can be found in
<url:distributions.txt:distributions.txt>
"probability rules" are how probabilities of multiple events are combined,
which is different for independent and dependent events, and if it's an
intersection or a union
We need a chart to depict probability rules, as they are different for "union"
and for "intersection" and conditional probability. I can't do an html chart in text-mode yet, so let me give first the listings for INDEPENDENT events
P(A union B) = P(A) + P(B) - P(B intersection A)
P(A intersection B) = P(A)*P(B)
P(A|B) = P(A)
example: intuitively, we say that two events E1 and E2 are independent if the probability of one event happening is in no way affected by the occurrence of the other event. So that P(E1|E2) = P(E1).
now, for DEPENDENT events:
P(A union B) = P(A)P(B|A)
P(A intersection B) = P(A)*P(B)
P(A|B) = P(A intersection B) / P(B)
source: http://www.mis.coventry.ac.uk/jtm/slides/19/sld19p2.pdf
(this uses the thought that we multiply the particular case by the number of
times that particular case will arise)
mutual exclusion: Two events E1 and E2 are said to be mutually exclusive if they do not have any common asmple points, i.e. the intersection of E1 and E2 is the null set.
example of mutual exclusion: If we toss 3 coins, let E1 be the event that there are two H's and let E2 be the event that there are at least 2 T's. These two events are mutually exclusive since there are no common sample points. On the other hand, if E2 is defined to be the event that there is at least one T then E1 and E2 are not mutually exclusive because they will have THH, HTH, HHT as common sample points.
The probability of an event is denoted by P(E).
If E1 and E2 are two events the probability of E1 or E2 or both happening is denoted by Probability(E1 UNION E2).
The probability of both E1 and E2 occurring at the same time is denoted by Probability(E1 INTERSECTION E2). The corresponding event is E1 INTERSECTION E2.
thm: P(E') = 1 - P(E)
P(E1 UNION E2) = P(E1) + P(E2) - P(E1 INTERSECTION E2) <= P(e1) + P(E2)
consider the experiment of tossing 4 coins. let E1 be the event that the number of Hs is even and let E2 be the event that there is at least on H. Then, E2 is the complemnt of TTTT. That is, the probablity of TTTT is 1/16, so P(E2) = 1 - 15/16 = 15/16. P(E1 INTERSECTION E2) = 7/16 since E1 INTERSECTION E2 = HHHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH. Thus P(E1|E2) = 7/16 / 15/16 = 7/16
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