probability definitions of statistics

7/27/2019 Probability definitions of statistics

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Probability


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Lets start with some definitions

Suppose you throw a pair of dice and add the numbers facing up (experiment)

When you repeat this experiment long enough you will find out that the set of allpossible outcomes is S = {2,3,4,,12}. S is called the sample space.

Sometimes you are interested in a subset of S. For example, you may want toknow, in the example above, the set of all possible even outcomes. We usually

refer to such subsets as events (E).In this case E = {2,4,6,8,10,12}.

Question: A coin is tossed three times and number of Heads are noted. What is thesample space for this experiment? What is the set of all outcomes where heads

come up at least twice?

S = {0,1,2,3}

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

E U E = S and E E =


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Another example

A pair of dice (one red, one green) is thrown and the number facing up on each die is

noted, let E be the event that the sum of the numbers is 4, and let F be the event that the

sum is an odd number.

a) The event F' is:

{(1, 1), (1, 3), (1, 5), (2, 2),(2, 4), (2, 6), (3, 1), (3, 3),

(3, 5), (4, 2), (4, 4), (4, 6),

(5, 1), (5, 3), (5, 5), (6, 2),

(6, 4), (6, 6)}

b) E F' is the event:

{(1, 3), (2, 2), (3, 1)}

c) E' U F is the event:

The sum of the numbers is any number other than 4


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Estimated probability

a) The frequencyof the event E [denoted as fr(E)] is the number of times the event E

occurs.

Example: Toss a coin 20 times. If heads comes up 13 times, then the frequency of the

event that heads comes up is fr(E) = 13.

b) Therelative frequencyorestimated probabilityof the event E is the fraction of

times E occurs: P(E) = fr(E) / N.

Referring to the example above, the estimated probability of the event that heads comes

up is 13/20=65%.

c) The number of times that the experiment is performed is called the sample size (N).

The experiment above was performed 20 times, so the sample size is N = 20.


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Estimated probability distribution

d) The collection of the estimated probabilities ofallthe outcomes is the estimated

probability distribution.

Example: If 10 rolls of a die resulted in the outcomes 2, 1, 4, 4, 5, 6, 1, 2, 2, 1, the

associated estimated probability distribution is the one shown in the following table.

Outcome 1 2 3 4 5 6

Probability 0.3 0.3 0 0.2 0.1 0.1


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Properties of probability distributions

Let S = {s1, s2, ... , sn} be a sample space and let P(s i) be the estimated probability of the

event {si}. Then:

0 P(si) 1

The estimated probability of each outcome is a number between 0 and 1.

P(s1) + P(s2) + .. + P(sn)=1

The estimated probabilities of all the outcomes add up to 1.

If E = {e1, e2, ..., er}, then P(E) = P(e1) + P(e2) + ... + P(er).

The estimated probability of an event E is the sum of the estimated probabilities of the

individual outcomes in E.


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Theoretical probability

Definition:The theoretical probability, orprobability, P(E), of an event E is the fractionof times we expectE to occur if we repeat the same experiment over and over.

Relationship with estimated probability:The estimated probability approaches thetheoretical probability as the number of trials gets larger and larger. Thus, estimatedprobability is an approximation, or estimate, of theoretical probability.

Determining theoretical probability:Theoretical probability is determined analytically, thatis, by using our knowledge about the nature of the experiment rather than through actualexperimentation. The best we can obtain through actual experimentation is an estimate ofthe theoretical probability (hence the term "estimated probability").

Example: Toss a fair coin and observe the uppermost side. Since we expect that heads isas likely to come up as tails, we conclude that the theoretical probability distribution is:P(H) = P(T) = 1/2.

Similarly, if you roll a fair dice: P(1)=P(2)= . . . = P(6) = 1/6.


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Probability distributions

Formal def ini t ion:A finite sample space is just any old finite set S = {s1, s2, ..., sn}. Aprobability distribution is an assignment of a number P(s i) to each outcome si in asample space S ={s1, s2, ... , sn} such that

0 P(si) 1The estimated probability of each outcome is a number between 0 and 1.

P(s1) + P(s2) + .. + P(sn)=1

The estimated probabilities of all the outcomes add up to 1.

A few more facts to note:

To get the probability of an event(recall that an event is a set of outcomes) all we dois add the probabilities of the individual outcomes.

If P(E) = 0, we call E an impossible event.


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An example

Fill in the missing probabilities, given that P(6) = 3P(1).

What is the probability that an even number will come up?

Outcome 1 2 3 4 5 6

Probability ? 0.125 0.125 0.125 0.125 ?


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Addition principle

a) If E and F are mutually exclusive events, then P(E U F) = P(E) + P(F).

Example: Cast two dice and add the numbers facing up. E: the sum is 5; F: the sum is

even. Since these are mutually exclusive events, the probability that the sum is either 5

or even is: P(E U F) = P(E) + P(F) = 1/9 + 1/2 = 11/18.

b) If E and F are NOT mutually exclusive events, then we must use the more general

formula: P(E U F) = P(E) + P(F) - P(E F).

Example: Cast a single dice and note the number facing up. E: the outcome is 1, 2 or 3;

so P(E) = 1/2; F: the outcome is even; P(E) = 1/2

E F = {2}; P(E F) = 1/6.

P(E U F)=P(E)+P(F)-P(E F) = 1/9 + 1/2 1/6 = 11/18.


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Other principles

P(E') = 1 - P(E)

P(S) = 1

P(O) = 1 P(S) = 0


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Conditional probability

Quick i l lust rat ion: Suppose you cast two dice; one red and one green.

Then the probability of getting "bulls eyes" (two ones) is 1/36.However, if, after casting the dice, you ascertain that the green die

shows a one (but know nothing about the red die), then there is a 1/6

change that both of them will be one. In other words, the probability of

"bulls eyes" changes if you have partial information, and we refer tothis (altered) probability as "conditional probability".


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Example

Used

cream Placebo TOTAL

Skin improved 800 600 1,400

No improvement 400 200 600

TOTAL 1,200 800 2,000

The table above shows fictitious trial results of a new acne cream.

The sample size (total number of people in the study) is: 2,000

The estimated probability that someone's skin improved (regardless of which skin

cream was used) is: 1,400/2,000 = 70%. (unconditional probability)

The experimental probability that someone's skin improved, given that they used the

new acne cream is: 800/1,200 = 66.7% (conditional probability)


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Formally

The probability of the event E, given the event F, and written P(E|F) (probability of E,given F)

If E and F are not mutually exclusive events, then the probability of E given F is

If all outcomes are equally likely, then we can also use the alternative formula

)(

)()|(

FP

FEPFEP

)(

)()|(

Ffr

FEfrFEP


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Another example

You have invested in Home-Clone Inc. stocks, as you suspect that the company's "Clone-

a-Sibling" kit will shortly be approved by the FDA. There is an 80% chance that FDA

approval will be given, and a 95% chance that the value of the stock you hold will double

if FDA approval is given. What is the probability that the value of the stock you hold will

double?

F: FDA approval, E: Price doubles

P(E|F) = 95%, P(F) = 80%, P(E) = ?

Using the formula above: 0.95 = 0.8 * P(E)

P(E) = 76%.


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Independent Events

If E and F are independent if any one of the following equations holds:

P(E|F) = P(E)

P(F|E) = P(F)

P(EF) = P(E) * P(F)

Example: You throw two fair dice, one green and one red, and observe the numbersuppermost.

A: the event that their sum is 7; P(E) = n(E)/36 = 6/36 = 1/6

B: the event that the red die shows an even number; P(F) = 1/2

P(AB) = P((1, 6), (3, 4), (5, 2)) = 3/36 = 1/12.

Since P(AB) = P(A) * P(B), A and B are independent.


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Bayes Theorem

Bayes' theorem gives us a way of calculating P(E|F) from a knowledge of P(F|E).

)'(*)'|()(*)|(

)(*)|()|(

FPFEPFPFEP

FPFEPEFP

Example: A manufacturer claims that its drug test will detect steroid use (that is, showpositive for an athlete who uses steroids) 95% of the time. Your friend on the rugby

team has just tested positive. What is the probability that he uses steroids is?

E = the event that an anabolic steroid detection test gives a positive result

F = the event that the athlete uses steroids

P(E|F) = 95%, P(F|E) = ?

To answer this, we need to know:

a) P(F) (the probability that an athlete on the team is using steroids)

b) P(E|F') (the probability of a false positive result).


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Example continues

Suppose, 15% of all steroid-free individuals also test positive [P(E|F') =15%]. Also,10% of the rugby team members use steroids [P(F)=10%]. Your friend on the rugby

team has just tested positive. The probability that he uses steroids is:

413.0

90.0*15.010.0*95.0

10.0*95.0)|(

EFP


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probability definitions of statistics

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