Chapter 5
Basic Concepts of Probability
Terminology
Usually, probabilities are descriptions of the likelihood of some event occurring (ranging from 0 to 1)The probability of two events occurring will be termed independent if knowledge of the occurrence of non-occurrence of the first event provide does not effect estimates of the probability that the second event will occur
Two events are termed mutually exclusive if the occurrence of one event precludes occurrence of the other event
A set of events are termed exhaustive if they embody all of the possible outcomes in some situation
Basic Laws of Probability
The additive law of probabilities: given a set of mutually exclusive events, the probability of occurrence of one event or another event is equal to the sum of their separate probabilitiesExample:
Place 100 marbles in a bag; 35 blue, 45 red, and 20 yellow.
P(blue)=.35 P(red)=.45 P(yellow)=.20
What is the probability of choosing either a red or a yellow marble from the bag?
P(red or yellow) = P(red)+P(yellow)
= .45+.20
= .65
The multiplicative law of probabilities: The probability of the joint occurrence of two or more independent events is the product of their individual probabilitiesExample:
Say that the probability that I am in my office at any given moment of the typical school day is .65
Also, say that the probability that someone is looking for me in my office at any given moment of the school day is .15
What is the probability that that during some particular moment, I am in my office and someone looks for me there?
P(in office and someone looks)
= P(in office) x P(someone looks)
= .65 x .15
= .0975
Question:
Say than Fred takes the car into work with a probability of .50, walks with a probability of .20, and takes public transit with a probability of .30
Barney, on the other hand, drives into work with a probability of .20, walks with a probability of .65, and takes public transit with a probability of .15
What is the probability that Fred walked or drove to work and Barney walked or took public transit to work, assuming Fred and Barney's behaviour to be independent?
Joint and Conditional Probabilities
The joint probability of two events A & B is the likelihood that both events will occur and is denoted as P(A,B)When the two events are independent, the joint probability simply follows the multiplicative rule
thus, P(A,B) = P(A) x P(B)
When they are not independent, it gets a little trickier ... but we won't worry about that for now
A conditional probability is the probability that some event (A) will occur, given that some other even (B) has occurred
denoted as P(A|B)
An Example: Drinking & Driving
Accident | No Accident | Total | |
Drinking | 7 | 23 | 30 |
Not Drinking | 6 | 64 | 70 |
Total | 13 | 87 | 100 |
P(Drinking) = 30/100 = 0.3000
P(Accident) = 13/100 = 0.1300
P(Drinking, Accident)
= P(Drinking) x P(Accident)
= 0.30 x 0.13
= 0.0390
P(Drinking | Accident)
= 7/13
= 0.5385
P(Accident | Drinking)
= 7/30
= 0.2333
Factorials!
We will soon discuss the concepts of permutations and combinationsPrior to that, it is necessary to understand another mathematical symbol, the symbol ! (read `factorial')
N! = (N) x (N-1) x (N-2) x ... x (1)
5! = 5 x 4 x 3 x 2 x 1 = 120
3! = 3 x 2 x 1 = 6
Note: 0! = 1
Permutations
If 3 people (p1, p2, & p3) entered a race, how many different finishing orders are possible?
p1, p2, p3 p1, p3, p2
p2, p1, p3 p2, p3, p1
p3, p1, p2 p3, p2, p1
each of these is called a permutation of the three people, taken three at a time
In permutation notation, this problem would be represented as and the answer could be solved using the following formula:
Another Permutation Example:
Say that 5 people entered the previous race (p1 thru p5), but only the first two get prizes. How many different orderings of those first two positions are possible?
p1, p2 p1, p3 p1, p4 p1, p5
p2, p1 p2, p3 p2, p4 p2, p5
p3, p1 p3, p2 p3, p4 p3, p5
p4, p1 p4, p2 p4, p3 p4, p5
p5, p1 p5, p2 p5, p3 p5, p4
Note: When doing permutations, order is important! Think of the word "permutations" as "orderings"
Combinations
Sometimes, we are not concerned about ordering but only in how many ways certain things can be combined into groupsFor example, let's say we again have five people and we want to form a team of two people. How many different teams of two people can we from our original five?
1&2 1&3 1&4 1&5 2&3
2&4 2&5 3&4 3&5 4&5
Other Examples:
Say you had seven people and you wanted to form a committee of four people who would work together (on equal footing) to solve some problem. How many different committees are possible?
What if the above committees were set up such that one person was to be president, the next vice president, the third treasurer, and the fourth secretary. Now how many committees are possible?
The Binomial Distribution
The binomial distribution occurs in situations in which each of a number of independent trials (termed Bernouli trials) results in one of two mutually exclusive outcomese.g., coin tosses
The mathematical description of the binomial distribution is the following:
where:
p(X) = The probability of X successes
N = The number of trials
p = The probability of success on any given trial
q = The probability of failure on any given trial (i.e., 1-p)
= The number of combinations of N things taken X at a time
Examples:
Suppose a batter (in baseball) gets a hit with a probability of 0.3, and gets out the rest of the time. What is the probability of that batter getting 0 hits in 10 at bats?
What is the probability of flipping a fair coin eight times and getting only two heads?
Plotting Binomial Distributions
Given that we can obtain probabilities for any value of X associated with some level of p, we can also use this probabilities to create a probability distributionFor example, if we toss a fair coin ten times, the following table represents the probabilities associated with the indicated outcomes:
Number Heads
|
Probability
|
0
|
.001
|
1
|
.010
|
2
|
.044
|
3
|
.117
|
4
|
.205
|
5
|
.246
|
6
|
.205
|
7
|
.117
|
8
|
.044
|
9
|
.010
|
10
|
.001
|
If these values were plotted as a distribution, they would look like:
Note 3 things; 1) these probabilities are discreet, 2) plotting probabilities, not frequency, 3) mathematically derived data, not empirically acquired
Mean of a binomial =
Variance of a binomial =
Standard Deviation =
Testing Hypotheses
Given all this, we can now ask questions like the following ...
Let's say a person is performing a true/false exam. How many questions out of 10 does a person have to get correct for us to reject the notion that they are just guessing?
Number Correct | Probability |
0 | .001 |
1 | .010 |
2 | .044 |
3 | .117 |
4 | .205 |
5 | .246 |
6 | .205 |
7 | .117 |
8 | .044 |
9 | .010 |
10 | .001 |
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