production date 2/20/00
Probability
The topic of probability provides you with a convenient bridge between descriptive
statistics and inferential statistics. Probabilities have both descriptive
and inferential functions.
When probabilities are used to describe a particular event, they are describing the likelihood of that event happening. For example, when a classmate states "I think the probability of quiz tomorrow is about 40%" they are describing what they think is the probability of that particular event. When the weatherman states that there is an 80 percent chance of rain tomorrow, they are describing what they think is the probability of that event.
When probabilities are used to test hypotheses, we can infer the truth about a hypothesis. A hypothesis is stated, data are
collected, and an inferential statistic is calculated. If the probability
of this inferential statistic is low, considering the hypothesis, the hypothesis
is rejected (considered false). If the probability of the statistic is not
low considering the stated hypothesis, then the hypothesis is not rejected.
All of this will become quite clear when we start conducting hypothesis tests.
Scientists can never be absolutely certain about most of their statements.
Statements and suppositions are usually based on samples
and not on populations. Scientists use probability to help them make correct
statements. Using probabilities, scientists communicate the degree of
certainty as to the truth of their conclusions. An example may help clear
up these points. Let's suppose that a friend asks you if their die is fair.
You could look at the die and see that it has six sides, all with different
spots (1-6). Would you tell your friend that their die is fair based on your
visual examination of the die? Obviously, you would not.
What would make you believe that this die is fair? You would roll
the die many times, and all the sides of the die would have to appear an equal number of times for the
die to be fair. To be absolutely sure that your friend's die is fair you
would need to roll it an infinite number of times. Of course, this is impossible.
How many times do you think would be reasonable to roll the die to answer
your friend's question? Would six times do? How about 12?
Let's suppose that you roll the die 120 times. Remember that the supposition
is that the die is fair. If it is, you expect that each side would
appear equally. You would expect the the table of values on the left:
Expected results from rolling a fair die 120 times.
|
Side
|
Frequency
|
|
1 spot
|
20
|
|
2 spot
|
20
|
|
3 spot
|
20
|
|
4 spot
|
20
|
|
5 spot
|
20
|
|
6 spot
|
20
|
|
Total
|
120
|
Found results from rolling a fair die 120 times.
|
Side
|
Frequency
|
|
1 spot
|
19
|
|
2 spot
|
21
|
|
3 spot
|
20
|
|
4 spot
|
19
|
|
5 spot
|
18
|
|
6 spot
|
23
|
|
Total
|
120
|
If you find a frequency distribution identical to the expected frequency
distribution, you would conclude that the die is fair. If you find a frequency
distribution dramatically different than the expected distribution, then
you would conclude that the die is not fair. What if the found distribution
was not identical to the expected distribution, but was just a little different?
What if you found the table shown on the right?
I hope that you think that the expected distribution and the found distribution
are not so different that the die might not be fair. The two distributions do
not have to be exactly alike. Since you are sampling a limited number of
throws (120), you would expect your results to be influenced by what scientists
call sampling fluctuations.
You have two different competing hypotheses in this problem.
1. The die is fair.
2. The die is biased.
You can not conclude which of these two hypotheses are absolutely true, you
can just determine which of the hypotheses is probably true based on the
results of your experiment. To make this decision, you need to study probability.
You will need to calculate the probability of your results. If the probability
of the found distribution is quite low given that the die is fair, then you
may reject the hypothesis that the die is fair, and consequentially conclude that the die is biased.
Simple probabilities are used to calculate the probability of a specific
event occurring. We often calculate simple probabilities without resorting
to a formula. Indeed, providing the formula often complicates matters for
those who do not appreciate the formula's utility. Try some
calculations with accompanying simulations on the following three separate pages.
- Coin flipping
- Die rolling
- Card Drawing
Usually these calculations are quite easy. Now for the formula.
The equations below provide the formulas for calculating simple probability.
The first one is written out, while the second is in symbol form.
![[Image]](Images/Chap9_pict0.jpeg)
Calculating the probability of getting a Jack from 1 draw of a well shuffled deck of
cards, using equation above, take the number of ways a specific Jack
can occur (4 - Jack of Hearts, Clubs, Spades, Diamonds) and divide by the
total number of possible events (52 cards) thus 4/52 = 0.0769.
Probability Tidbits
The probability of an event occurring will always
be between 0.0 and 1.0.
When we looked at the areas under the z distribution in Chapter 6, we calculated
probabilities that these z values would occur. For example, given a normal
distribution, the probability that a score picked at random would have a z
value between -1 and +1 is 0.6826.
Additive Law of Probability
Determining the probability of two events occurring,
use the additive law of probability. For example if asked "What
is the probability of drawing a 4 or a 7 from a shuffled deck of cards?",
Eight different cards would satisfy this condition, 4 (4s) or
4 (7s). So the probability of this event is 8/52.
The equation below properly states the additive law of probability.
![[Image]](Images/Chap9_pict1.jpeg)
The probability of either of two events A or B occurring is the probability
of event A plus the probability of event B minus the probability that both
A and B will occur.
Using the additive law of probability you can find the probability that in
one roll of a die, you will obtain either a one-spot or a six-spot. The
probability of obtaining a one-spot is 1/6. The probability of obtaining
a six-spot is also 1/6. The probability of rolling a die and getting a side
that has both a one-spot with a six-spot is 0. There is no side on a die
that has both these events. So substituting these values into the equation
gives the following result:
![[Image]](Images/Chap9_pict2.jpeg)
Finding the probability of drawing a 4 of hearts or a 6 or any suit using
the additive law of probability would give the following:
There is only a single 4 of hearts, there are 4 sixes in the deck and there
isn't a single card that is both the 4 of hearts and a six of any suit.
Now using the additive law of probability, you can find the probability of
drawing either a king or any club from a deck of shuffled cards. The equation
would be completed like this:
![[Image]](Images/Chap9_pict4.jpeg)
There are 4 kings, 13 clubs, and obviously one card is both a king and a
club. You don't want to count that card twice, so you must subtract one of
it's occurrences away to obtain the result.
Multiplicative Law of Probability
The multiplicative law of probability is used to calculate the probability
that two events will occur in sequence. Remember that the additive law was
used to find the probability that one of two different events would occur
on a single trial. These two probabilities have key words to look for. In
the additive law, you will always see something like the following: Find
the probability that event (A) or event (B) will
occur. The word or is key. With the
multiplicative law, the word and is key. Here is
a typical problem requiring the multiplicative law: What is the probability
of drawing a 4 and then a 7 from a shuffled deck
of cards.
The multiplicative law states that the probability of the occurrence of event
A and then event B, and then event C ... and so on is the product of the
separate simple probabilities of those events. The following equation expresses the
multiplicative law of probability's formula.
![[Image]](Images/Chap9_pict5.jpeg)
So in finding the probability of drawing a 4 and then a 7 from a well shuffled
deck of cards, this law would state that we need to multiply those separate
probabilities together. Completing the equation above gives:
![[Image]](Images/Chap9_pict6.jpeg)
Given a well shuffled deck of cards, what is the probability of drawing a
Jack of Hearts, Queen of Hearts, King of Hearts, Ace of Hearts, and 10 of
Hearts?
![[Image]](Images/Chap9_pict7.jpeg)
or in scientific notation this might be written as 2.6301667E-9. Remember that he first
part of this notation are the digits other than zeros. The second part (E
- 9) tells us how many places the decimal written in the first part needs
to be moved to obtain the correct answer. In this case we need to move the
decimal place back 9 places. In any case, given a well shuffled deck of cards,
obtaining this assortment of cards, drawing one at a time and returning it
to the deck would be highly unlikely (it has an exceedingly low probability).
So far, we have assumed that we were sampling with
replacement (we put our cards back in the deck after we noted what
they were). When an element is not returned, but kept, we are
sampling without replacement. Often this changes
our simple probability calculations.
What is the probability of drawing a 4 and then a 7 from a well shuffled
deck of cards, if you keep the first card? The multiplicative law would say
that there are four 4s so if you draw a 4 first, it's probability is 4/52.
If you keep the 4 and then attempt to draw a 7 from the deck, there are still
four 7s left in the deck, but the total number of cards in the deck are now
51. You have one in your hand, and 51 left in the shuffled deck. The probability
of this two card hand is thus:
![[Image]](Images/Chap9_pict8.jpeg)
Now find the probability of drawing two 10s from a well shuffled deck of
cards assuming that the first 10 is kept.
![[Image]](Images/Chap9_pict9.jpeg)
Notice that both the numerator and denominator of the second fraction are
changed. There are only three 10s left if one is kept, and 51 of
the 52 original cards remain.
Sometimes it is very difficult to count the specific number of events or
the total number of events in a probability calculation. We have been dealing
with cards and die, precisely because there they are easy to calculate. In
order to calculate the number of ways events can occur in any situation,
we need to study permutations and combinations.
Permutations
Permutations of 4 things 2 at a time.
|
AB
|
BA
|
|
AC
|
CA
|
|
AD
|
DA
|
|
BC
|
CB
|
|
BD
|
DB
|
|
CD
|
DC
|
A permutation is an ordered sequence of objects or events. The keyword
is ordered. Order matters. Suppose you have four
different objects (A,B,C,D). How many different ways can they be put together
two at a time where their order makes a difference (AB is different than
BA). This is a permutations problem. With such a limited number of objects,
we don't have to resort to the formula initially. Just figure this
one out. All the different permutations are shown in the table on the left.
There are 12 possible permutations. The number of permutations of N events
or objects takes r at a time is given by the equation directly below.
![[Image]](Images/Chap9_pict10.jpeg)
The first thing in this equation that needs explained is most likely
the exclamation point (!). This is called a
factorial. Factorials are found by subtracting one
from the beginning number and multiplying then subtracting another one and
multiplying until one gets to the number one. For example 5! is equal to
5*4*3*2*1 = 120. By definition 0! = 1. Now, solving the problem above using
the permutations formula. How many permutations are possible with 4 object
taking 2 of them at a time? N = 4, and r = 2. The equation would give:
![[Image]](Images/Chap9_pict11.jpeg)
This is the same answer as we achieved when we wrote them all out above.
Try a more complicated example. What if all 3-letter orders were meaningful
words? Further suppose that we never wanted to use a letter more than once
in any of our words. How many different 3-letter words could be form using
our alphabet and using no letter more than once? This is certainly a permutations
problem, and one where you wouldn't want to take a piece of scratch paper
and try to figure the answer out by trial-and-error. The permutations formula
quickly gives us the answer.
![[Image]](Images/Chap9_pict12.jpeg)
There would be fifteen thousand six hundred possible words using the 26 letters
of the English alphabet.
Combinations
Combinations of 4 things 2 at a time.
|
AB
|
|
AC
|
|
AD
|
|
BC
|
|
BD
|
|
CD
|
A combination is a sequence of objects or events where order is not important.
Thus, AB and BA are treated as if they are the same thing. If you have four
objects (A,B,C,D) there are only six combinations when two of these objects
are put together at a time. The six combinations are shown in the table on the left.
The combinations formula given below gives us the same answer.
![[Image]](Images/Chap9_pict13.jpeg)
Filling in the combination equation for the number of combinations of 4 objects taken
2 at a time yields:
![[Image]](Images/Chap9_pict14.jpeg)
Work another more difficult problem. What is the
probability of being dealt a five card poker hand containing all
diamonds?
To calculate the probability of such an event, you will need to
know the number of five-card poker hands that are possible with
all diamonds, and the number of five-card hands that are possible
in the deck of cards. Remember the simple probability formula:
![[Image]](Images/Chap9_pict15.jpeg)
Are these permutation or combination problems? In playing poker, does it
matter in which order the cards were drawn? Obviously order is not important,
so we are dealing with combinations.
To calculate the number of possible 5 card diamond hands (Ns in the above
equation) we need to calculate the number of different combinations that
are possible using the 13 diamonds 5 cards at a time.
![[Image]](Images/Chap9_pict16.jpeg)
There are one thousand two hundred and eighty seven different all diamond
5-card hands possible.
When calculating the total number of possible 5-card poker hands,
there are 52 cards available. To calculate the number of combinations
of 52 cards using 5 at a time that are possible (Nt), the following combinations
formula must be solved.
![[Image]](Images/Chap9_pict17.jpeg)
Finally, calculating the probability of the 5-card diamond hand being dealt from a shuffled deck of cards combine the two combinations solutions
using the simple probability formula.
![[Image]](Images/Chap9_pict18.jpeg)
Thus 5 in every 10 thousand hands dealt will be 5-card diamond hands.
So far, we have discussed what most textbooks consider elementary probability
topics. We have discussed simple probability, the additive law, and the
multiplicative law [p(A), p(A or B), p(A and B)]. To apply probability notions
to inferential statistical decisions, we need to study and understand,
conditional probability, joint and marginal probabilities, and finally, the
binomial probability distribution. With these, we can understand the more
complex probability distributions used in inferential decisions.
Conditional Probability
Conditional probabilities are calculated when we need to know the likelihood
of event A happening given that event B has already happened. We say that
event A is conditional on event B. Conditional probabilities don't have a
keyword, they have a key-symbol (|). Conditional probabilities are written
p(A|B), which can be read "The probability of A given B". Some simple problems
are in order first. Then the discussion will turn to joint and marginal probabilities
before returning to conditional probabilities.
Find the probability of drawing a 4 from a shuffled deck of cards given that
you have already drawn a 7 from the deck. If you have drawn the 7 then only
51 of the 52 cards are available so the probability would be calculated like
this:
p(4 | 7) = 4\51 = 0.0784.
What is the probability of drawing a second 10 from a deck of cards given
that you have drawn the first 10.
p(10 | 10) = 3/51 = 0.0588
Drawing the first 10 affected both the numerator and denominator of the
probability. With one 10 drawn, there are only three 10s left and only 51
total cards.
Joint and Marginal Probabilities
Joint and Marginal probabilities are found when you are working with problems
which have two variables or dimensions of concern. Suppose you are a psychologist
interested in mental health diagnosis. You interview 100 psychiatric patients
and give them the Rorschach test (inkblots). One of the variables that might
interest you is whether the patient reports movement when they describe what
they see while looking at the inkblot. This variable will only have two levels.
Either the patient interprets movement or they fail to interpret motion.
The second variable that might interest you is whether they are classified
as psychotic or neurotic. This second variable only has two different values.
This is a very simple case with two variables of interest, each with only two different values. When experimental results can only have two different outcomes, they are called
Bernoulli trials. The figure below illustrates what the interview results
might be.
![[Image]](Images/Chap9_pict20.jpeg)
The figure directly above shows that out of the 100 interviewed patients, 50 patients
were both Neurotic and interpreted motion. Twenty five patients were classified
as Neurotic and did not interpret motion when they discussed the inkblots.
Ten patients were psychotic and interpreted motion, while 15 patients were
psychotic and failed to interpret motion. You will also notice that the totals
for each row and column are printed in the margins of the figure. For example,
75 patients were classified as neurotic, and 25 as psychotic. Likewise, 60
patients interpreted motion, while 40 did not. In the lower right of the
figure both the column and row total is indicated as 100.
In the figure below, probabilities are calculated using the counts that were found
in the figure above. The probabilities that are found in the lower triangle of
each box are called joint probabilities. For example, the .5 in the upper
left box is the probability of jointly being neurotic and interpreting motion.
It is simply found by dividing the number of patients who are jointly neurotic
and interpret motion (50) by the total number of patients. In each margin
you will find the marginal probabilities. For example .6 in the upper left
margin indicates that 60 percent of the patients interpreted motion in this
sample.
![[Image]](Images/Page1_pict0.jpeg)
When we have problems with more than one variable of interest then there
are some quite interesting conditional probabilities that can be calculated.
Suppose one wishes to calculate the probability of interpreting motion given
that the patient is psychotic. Obviously, all the neurotic patients are of
no concern in this calculation, so they are eliminated. The next figure illustrates
this. The p(Interpreting motion | psychotic) is now 10/25 = .4. There are
10 of the 25 psychotic who interpreted motion. These conditional probabilities
can also be calculated from the joint and marginal probabilities without
resorting to the counts. For example p(Interpreting motion | psychotic) is
.1/.25 = .4.
![[Image]](Images/Page1_pict1.jpeg)
Now calculate the probability of being psychotic given that the patient
interpreted motion. Obviously, all the patients who failed to interpret
motion are not needed. Again, the following figure illustrates the counts, and joint and marginal
probabilities that can be used to calculate this conditional probability.
![[Image]](Images/shock_pict2.jpeg)
One can calculate that the p(Psychotic | Interpret Motion) is equal to 10/60
= .16. Or using the joint and marginal probabilities it is .1/.6 = .16. The following equation illustrates how conditional probabilities can be calculated from known
joint and marginal probabilities.
![[Image]](Images/shock_pict3.jpeg)
Where p(AB) is the joint probability of event A happening with event B.
The binomial probability distribution arises in any situation where there
are: (1) only two possible outcomes, (2) the number of trials or observations is fixed, (3) all the observations are independent, and (4) the probability of a success (p) is identical for each observation. Here is the first instance in this environment
where the same symbol is going to stand for two different things. In the
binomial situation the letter p stands for the
probability of a success. Before it has just stood for the word probability.
In the binomial situation q is used to designate
the probability of a failure. Success and failure are difficult words to
define at times within these problems. Just remember that there are only
two possible outcomes in these situations. One situation might be coded with
a one (1) while the other is coded with a zero (0). For example, you might
answer a question on a test correctly (pass) and receive one point, or answer
the question incorrectly (fail) and receive zero points. P would indicate
the probability of getting a one and q would indicate the probability of
getting a zero. On any event, p + q must always equal 1. You must do one
or the other if there are only two possible outcomes. As has been our custom,
we work a problem before presenting the formulas.
Suppose we are working at a ski lodge, and we find it necessary to calculate the
probability that exactly two of five persons in a party will break their
leg on the slopes. One of the first things we need to do, is calculate how
many ways 2 of five persons can break a leg. The figure below shows all the possible
ways. Note that B means the person broke their leg, while NB means they didn't (No Break).
![[Image]](Images/shock_pict4.jpeg)
Note that there are 10 different ways (combinations) that two people out
of five could break their leg. The combinations formula would also calculate this number.
Now let's calculate the probability of any one of these 10 combinations.
Let's try the first one. What is the probability that person 1 will break
their leg and then person two break their leg,
and then person three not break their leg,
and then person for fail to break their leg,
and then person five not break their leg? Which
law would you use to calculate this?
Hint: All the ands should provide a clue.
Of course, the multiplicative law is the correct answer. Note that if p is the probability of a
success (breaking your leg in this case), that there are two breaks in every
combination. If we multiply p * p we always have p squared or p2. Also if
q is the probability of a failure, or not breaking a leg, in each combination
there are exactly 3 failures. If we multiple q together three times we have
cubed it (q3). Thus 10p2q3 will work.
This is exactly what the formula for
the binomial probability calculates. The two equations below are the binomial
probability formulas with the second just substituting the factorials for the
combination found in the first. The binomial formulas may be used for any
binomial probability calculation.
![[Image]](Images/shock_pict5.jpeg)
Suppose that the probability of breaking your leg is 0.20.
Therefore the probability of not breaking your leg is 0.80. What is the
probability that exactly 2 out of 5 people will break their leg. N = 5, r
= 2, p = .2, q = .8. Completing the equation above yields:
![[Image]](Images/shock_pict6.jpeg)
Using the binomial probability equation find answers to the next three problems using hand calculations. In the very next section Statlets will be used to calculate the answers to these questions. Work the hand calculations before proceeding to the next section.
1. Find the probability that in 10 rolls of a fair die, you will get
exactly 6 three-spots.
2. What is the probability of obtaining exactly 4 heads in 6 flips
of a fair coin?
3. What is the probability of obtaining 4 or more heads in 6 flips
of a fair coin?
In the previous chapter you used Statlets Plot/Probability Distributions to draw the PDF and CDF functions for the binomial distribution. Here you will learn how to use the same function to calculate probabilities in the binomial distribution.
Working the first problem above, the question asks for the probability that in
10 rolls of a fair die that exactly 6 three-spots will occur. First start Statlets,
and select the Plot/Probability Distributions menus. In the Input tab, select
the Binomial distribution. Click the PDF tab, and using the Options button, set
the characteristics for this distribution. Specifically, set the Event Probability
(the p value) to the probability of getting a three-spot with a fair die. This
p value is 1/6 or .1666667. Next set the number of trials to 10 as stated in the
problem. Your Options button input should look like the figure directly below.
When the PDF graph is redrawn, the binomial distribution with these characteristics should look like the following figure.
Next click the Tail Areas tab and then the associated Options button. In the Options input, type in the number 6 for the number of three-spots. This Options' input should look like the following figure.
After clicking OK, the following Tail Areas output is provided.
The probability we are interested in for this problem is the one given under the Probability Mass (=) heading. Here Statlets is calculating that the probability of getting exactly 6 three-spots in 10 rolls of a fair die is 0.002171. Quite a small probability. Under the heading Lower Tail Area (<), the probability of getting zero to 5 three-spots is calculated. Under the heading Upper Tail Area (>), the probability of getting 7 through 10 three-spots is calculated. If one wanted to calculate the probability of getting 6 or more three spots, one would need to add the probabilities given under the Probability Mass (=) heading and the Upper Tail Area (>) headings.
For the second problem, click the PDF tab, and then using the Options button change the Event Probability to .5 (the probability of obtaining a head with a fair coin), and the Number of Trials to 6.0 as indicated in the second problem. The PDF graph should look like the following.
Next click the Tail Areas tab, and using the associated Options button, change the Critical Values entry to 4.0. The results as shown in the figure below should indicate that the probability of obtaining exactly 4 heads in six flips of a fair coin is 0.234375.
Finally, for the last problem, all you need to do to find the probability of obtaining four or more heads from flipping a fair coin is to add the probability of obtaining the four heads (0.234375) with the Upper Tail Probability (0.109374). Calculating the probability of obtaining four or more heads from six flips of a fair coin to be 0.343749.
At State University, the probability that students will complete their weekend
reading assignments on time is .77. What is the probability that for 10 or more
of the 18 weekends a student will complete the reading assignment. Using Statlets'
Plot/Probability Distributions procedure, enter the characteristics of this distribution,
and then using the Tail Areas tab, calculate the percentage requested. If your
instructor requests, submit this project
report.
Using the binomial distribution characteristics in Project 15, use the Critical Values tab, to calculate the critical value in this distribution that would have 10% of the scores below it. If your instructor requests, submit the project 16 report.
Binomial problems with a large N and both p and q equal
to .5, look very much like a normal distributions.
One can calculate the mean and standard deviation of large binomial distributions
with the following two equations.
![[Image]](Images/shock_pict7.jpeg)
Think about how you might decide if a coin was fair given that you flipped
it 20 times. How many heads would you expect to get? How many heads would
it take until you thought that the coin was weighted in some way?
Draw a series of PDF graphs for the binomial distribution where the Number of
Trials remains constant at 100, but the Event Probability changes using the following
values: 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9. How would you describe the
change in the skew of the binomial distribution as the Event Probability changes?
This link
allows you to take a computer scored end-of-chapter test. If your instructor requests
to see the results of this examination, you can either copy and e-mail or print
the feedback you will receive immediately after taking the test.
Please
send a report indicating your understanding of this chapter to your instructor.
You will need to know both your and your instructor's e-mail addresses.
