Harold Nelson
2/5/2019
Probability is always based on an experiment. The set of all possible outcomes of an experiment is known as a sample space.
The individual members of the sample space are described as elementary outcomes. The probability of an elementary outcome is best described as the fraction of the time that the elementary outcome will occur when the experiment is performed a large number of times.
The sum of all of the probabilities of the elementary outcomes must be 1.0.
An event is a subset of a sample space. The probability of an event is what we expect its relative frequency to approach as we run the experiment a large number of times. The probability of an event is the sum of the probabilities of its elementary outcomes.
Probabilities are always numbers in the range \(0\) to \(1\).
If the elementary outcomes of an experiment are discrete and all equally likely, the theoretical probability of an event is defined as
\[\frac{\text{Number of elementary outcomes in the event}}{\text{Total number of elementary outcomes}}\]
As an example, consider the probability of getting a head when you flip a fair coin. There are two elementary outcomes of this experiment, a head, or a tail. Since the coin is fair, the outcomes are equally likely. There are two possible outcomes and only one is in our event. So, the probability is \(1/2\). In everyday language, you may hear “fifty-fifty.” To stick with the language of probability we say that the probability is \(.5\).
Using equally likely elementary outcomes with discrete experiments is clearly advantageous. Consider the probability of getting a sum of seven dots when rolling a pair of dice. If we think of the elementary outcomes of this experiment as the numbers between two and twelve, the outcomes are not equally likely. There is only one way to get a sum of two and only one way to get sum of twelve, but there are many ways to get a sum of seven.
Create a table showing the possible sums of dots showing when you throw a pair of dice.
Are the elementary outcomes in this table equally likley?
What is the probability of each elementary outcome?
How many elementary outcomes are in the event “7”?
dice = matrix(rep(0,36),nrow=6)
rownames(dice) = 1:6
for( i in 1:6){
for (j in 1:6){
dice[i,j] = i + j
}
}
colnames(dice) = 1:6
dice## 1 2 3 4 5 6
## 1 2 3 4 5 6 7
## 2 3 4 5 6 7 8
## 3 4 5 6 7 8 9
## 4 5 6 7 8 9 10
## 5 6 7 8 9 10 11
## 6 7 8 9 10 11 12There are 36 possible equally likely elemementary outcomes when we consider the orderd pairs of outcomes describing what happened to die 1 and what happened to die 2. These are \((2,5),(5,2),(3,4),(4,3),(6,1) \text{ and } (1,6)\). Since there are six of them, the probabiliy of getting a seven is \(6/36\), or \(1/6\).
This is really just noting that you can describe some empirical facts using the language of probability. An example or two should help.
Based on the Current Population Survey, about 60% of the US population over the age of 16 meets the definition of participating in the labor force. We can say that if a member of the population over the age of 16 were picked at random, the probability of that person being a labor force participant is about .6.
Based on data from the CDC, there were 2,515,458 deaths in the US in 2010. Of those the primary cause of death was heart disease in 596,577 cases. We can say that if we were to pick a 2010 death certificate at random the probability of the death having a primary cause of heart disease is .237.
This is really just using the language of probability to express a personal degree of belief. Here are some examples based on my own personal experiences.
I believe that the probability of having at least one class cancelled during a winter term because of snow is about .5.
I believe that the probability of my laptop having a hardware failure during the next 6 months is less than .05.
There are several rules for constructing the probabilities of events defined in terms of other events.
\[P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)\] Example Draw a single card from a 52 card deck. What is the probability of a heart or an ace?
\[P(\text{Heart or Ace})=P(\text{Heart})+P(\text{Ace})-P(\text{Heart and Ace})\]
\[P(\text{Heart or Ace})=\frac{13}{52}+\frac{4}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}\]
The subtraction of the third term is necessary because the first two terms would have counted the ace of hearts twice.
If the two events are mutually exclsuive (disjoint), the third term may be omitted since its value is zero. \[P(A\text{ or }B)=P(A)+P(B)\] Example Draw a single card from a 52 card deck. What is the probability of a heart or a spade?
\[P(\text{Heart or Ace})=P(\text{Heart})+P(\text{Spade})\]
\[P(\text{Heart or Spade})=\frac{13}{52}+\frac{13}{52}=\frac{26}{52}=\frac{1}{2}\] The simpler version is applicable because no card can be both a heart and a spade.We say that these two events are disjoint or mutually exclusive.
If A is an event, either it occcurs or does not occur whenever the experiment we are thinking of occurs. If we think of events as subsets of a sample space, then every outcome is in the subset A or not in the subset A. The set of all outcomes not in the event A is called the complement of A, sometimes denoted \(A^{c}\). No outcome is both in A and in \(A^{c}\) (mutually exclusive). Also, every outcome is in one of these (exhaustive). In set theoretic terms \(A \cup A^{c} = \text{Sample Space}\).
The practical consequence is that \[P(\text{A}) + P(\text{not A}) = 1.0\]
Example What is the probability of not getting a heart when you draw a single card from a 52 card deck?
\[P(\text{not Heart}) = 1 - P(\text{Heart}) = 1 -\frac{13}{52} = \frac{3}{4}\]
\[P(\text{A and B}) = P(\text{A})*P(\text{B given that A has occured}).\] The phrasing in the second expression is generally replaced with a single vertical bar, ‘|’ between the two event names. The law is then written as \[P(\text{A and B}) = P(\text{A})*P(\text{B|A}).\]
Example What is the probability of drawing two hearts if you draw two cards from a 52 card deck without replacement. There are two events involved, getting a heart on the first draw (H1) and getting a heart on the second draw (H2).
\[P(\text{H1 and H2}) = P(\text{H1})*P(\text{H1|H2})\]
\[P(\text{H1 and H2}) = \frac{13}{52}*\frac{12}{51}=\frac{3}{51}=\frac{1}{17}\]
\[P(\text{A and B}) = P(\text{A})*P(\text{B}).\] If the probability that B will occur does not depend on whether A has occurred or not, we say that the two events are independent. In this case, we use the simple probability rather than the conditional probability.
Example What is the probability of drawing two hearts if you draw two cards from a 52 card deck with replacement. There are two events involved, getting a heart on the first draw (H1) and getting a heart on the second draw (H2). However, when we include the stipulation “with replacement,” the condition of the deck on the second draw is the same as it was on the first draw.
\[P(\text{H1 and H2}) = \frac{13}{52}*\frac{13}{52}=\frac{1}{4} *\frac{1}{4}=\frac{1}{16}\]
In the description of the multiplication law above, we have assumed that A occurs first and may (or may not) influence the probability that a second event, B occurs. We imagine that an observer, having observed A, obtains an improved estimate of the probability that B will occur. In some circumstances, it makes sense to reverse our thinking and become more like a detective. If B has been observed, what is the probability that A occurred?
Write two equivalent expressions for \(P(\text{A and B}).\) \[P(\text{A and B}) = P(\text{A})*P(\text{B|A}) = P(\text{B}) *P(\text{A|B}).\]
Now solve the equation relating the second and third equations for \(P(\text{A|B})\).
\[P(\text{A})*P(\text{B|A}) = P(\text{B}) *P(\text{A|B})\]
\[P(\text{A|B}) = \frac{P(\text{A})*P(\text{B|A})}{P(\text{B})}\]
In many cases, we don’t know the probability of B directly. However we can construct it. If B occured, either A occurred and then B occured or; A did not occur and then B occurred anyway. These two possibilities are mutually exclusive and exhaustive. Therefore we can write
\[P(\text{B}) = P(\text{A}) * P(\text{B|A})+P(\text{ not A}) *P(\text{B| not A})\]
Example Return to the problem of drawing two hearts from a 52 card deck without replacement. Suppose the second card was a heart. What is the probability that the first card was also a heart?
\[P(\text{H1|H2}) = \frac{P(\text{H1})*P(\text{H2|H1})}{P(\text{H2})}\]
Now rewrite the denominator
\[P(\text{H1|H2}) = \frac{P(\text{H1})*P(\text{H2|H1})}{P(\text{H1})*P(\text{H2|H1})+P(\text{not H1})*P(\text{H2 | not H1})}\]
Now let’s insert the numerical values.
\[P(\text{H1|H2}) = \frac{\frac{13}{52}*\frac{12}{51}}{\frac{13}{52}*\frac{12}{51}+\frac{39}{52}*\frac{13}{51}}\]
Your most likely use of probability concepts will be in interpreting a contingency table. Such a table is a display of the outcomes in terms of two categorical variables when taking a random sample from a population.
In a population of 100 people, 60 are men and 40 are women. 40 of the men smoke and 10 of the women smoke. Display the 4 cell table describing this data. Label the columns with Men/Women and the rows with Smoker/Non-Smoker.
## Men Women
## Smoker 40 10
## Non-Smoker 20 30Calulate the following.