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Relative Frequency Definition of ProbabilityThe relative frequency definition of probability says: If an experiment is repeated, the probability of an event (a specified outcome of the experiment) is the relative frequency of occurrence of that event in a large number of repetitions of the experiment. In this definition 'large number' is vague and will not be made precise here. The following links use the Virtual Laboratories in Probability and Statistics by Professor Kyle Siegrist of the University of Alabama at Huntsville. Each link opens a page that illustrates the relative frequency definition of probability for a different situation.
Sample Space Definition of Probability
|
| 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 | 12 |
- From
the table shown above P[A]=(Number of red
sums)/36=30/36, P[B]=(Number of underlined
sums)/36=18/36, and P[C]=(Number of bold italicized
sums)/36=5/36. P[Not A]=1-P[A]=1-(30/36)=6/36,
P[Not B]=1-P[B]=1-(18/36)=18/36, P[Not C]=1-P[C]=1-(5/36)=31/36.
- From
the table above P[A or B]=(Number of red or underlined (or both) sums)/36=33/36, P[A or C]=(Number of red or
bold italicized sums)/36=31/36, and P[B or C]=(Number of
underlined or bold italicized sums)/36=21/36.
- From the table above P[A and B]=(Number of red and underlined sums)/36=15/36, P[A and C]=(Number of red and bold italicized sums)/36=4/36, and P[B and C]=(Number of underlined and bold italicized sums)/36=2/36.
and C=Ace. Find P[A], P[B], P[C], P[Not A], P[Not B], P[Not C], P[A or B],
P[A or C], P[B or C], P[A and B], P[A and C], P[B and C].
- The
deck has 26 black cards, so P[A]=26/52, it has 13
spades, so P[B]=13/52, and it has 4 aces, so P[C]=4/52.
- In
the deck 13 cards are both black and spades, so P[A and
B]=13/52, 2 cards are both black
and aces, so P[A and C]=2/52, and one card is both a spade and an ace, so P[B and C]=1/52.
- In
an ordinary deck 26 cards are either black or a spade
(or both) so P[A or B]=26/52, 16 cards are either black
or an ace (or both), so P[A or C]=28/52, and 16 cards
are either a spade or an ace (or both), so P[B or
C]=16/52.
balls numbered 1 and 2 are green and balls numbered 3 and 4 are red. List all of the sample points in the experiment. What probability should be assigned to each of the sample points?
Conditional Probability and Independence
- Rules for Conditional Probability
- P[A|B] = P[A and B] / P[B]
- or P[A and B] = P[A | B] P[B] or P[A and B] = P[B | A] P[A]
- P[A|B] = P[A and B] / P[B]
- Conditional
Probability Calculations
- Consider drawing two cards, one after another without
replacement, from an ordinary deck. On drawing the first
card the sample space consists of 52 cards. However, once
the first card is drawn, the sample space for the second draw
consists of the 51 cards that remain. Then, for example, the
probability that the 2nd card is a heart given that the first card
was a heart is 12/51, while the probability that the 2nd card is a
heart given that the first card was not a heart is 13/51.
These probabilities are called conditional probabilities.
They are denoted by
P[2nd is heart | 1st is heart] and P[2nd is heart | 1st is not heart].
The vertical symbol | means given.
- A
ball is selected from a container that holds 6 red and
10 green balls. Then a second ball is selected
from the container without replacing the first ball.
The probability that the second ball is green given that
the first ball is green is 9/15 while the probability
that the second ball is green given that the first ball
is red is 10/15. These probabilities can be
written as
P[2nd ball green | 1st ball green]=9/15 and P[2nd ball green | 1st ball red]=10/15
- A
population is classified by gender and political party
affiliation. The membership of the population is
shown in the following table with gender along the top
and political party along the left side of the table.
The interior of the table givens the number of people in
each classification.
Female Male Democrat 45 50 Republican 60 44 Independent 10 15 A person is selected at random from this population. The following ordinary probabilities can be found directly from the table: P[Democrat]=95/224, P[Female]=115/224, P[Democrat and Female]=45/224. Also, the following conditional probabilites can be found directly from the table by working with the appropriate row or column of the table. P[Democrat | Female]= 45/115, P[Female | Democrat]= 45/95, and P[Independent | Male]= 15/109. From this example, notice that the following are true:
(1) P[Female | Democrat] = P[Female and Democrat]/P[Democrat]
(2) P[Independent and Male] = P[Independent | Male] P[Male].
The two rules, P[A | B] = P[A and B]/P[B] (for P[B] unequal to 0) and P[A and B] = P[A | B] P[B] are true for any two events, A and B.
- Consider drawing two cards, one after another without
replacement, from an ordinary deck. On drawing the first
card the sample space consists of 52 cards. However, once
the first card is drawn, the sample space for the second draw
consists of the 51 cards that remain. Then, for example, the
probability that the 2nd card is a heart given that the first card
was a heart is 12/51, while the probability that the 2nd card is a
heart given that the first card was not a heart is 13/51.
These probabilities are called conditional probabilities.
They are denoted by
- Independence
of Events: Events A and B are independent if and only if P[A |
B] = P[A]
- The definition of independence can be stated as A and B are
independent if and only if the occurrence of one of them, say
B, does not affect the occurrence of the other, A.
- Since P[A and B] = P[A | B] P[B], and since for independent
events, P[A | B]=P[A],
P[A and B] = P[A] P[B]
- In tossing a fair coin twice, the probability of event A,
getting heads on the first toss is 1/2. The probability
of event B, getting heads on the second toss is also 1/2.
The probability of event A and B, getting heads on the first
and second toss is 1/4. In this case P[A and B] = P[A]
P[B], so A and B are independent events.
- In drawing two balls without replacement from a container
that holds 6 red and 10 green balls. The probability of
event A, that the first ball picked is red, is 6/16. The
probability of event B, that the second ball picked is red, is
also 6/16. This second probability may seem surprising
but it can be shown to be true. The probability of event
A and B, that both balls are red, is (6/16)(5/15). In
this case P[A and B] is not equal to P[A] P[B], that is A
and B are not independent events.
- If the engines on a jet airplane are identical in their
ability to operate throughout a particular flight, if the
probability that any one of the engines works properly for a
particular flight is 0.97, if the engines operate
independently of one another, and if the plane can make this
particular flight safely if at least one engine continues
working, what is the probability that a jet plane with four of
these engines is able to make this flight safely?
P[Makes flight safely] = 1 - P[Doesn't make flight safely] = 1 - P[all engines fail] = 1 - 0.034 = 0.99999919 The key step was finding the probability that all engines fail. This was found by multiplying the probability that any one of the engines fail together 4 times--this is allowed by the assumption of independence for the engines.
- The definition of independence can be stated as A and B are
independent if and only if the occurrence of one of them, say
B, does not affect the occurrence of the other, A.
Random Variables
- Definition--A random variable is a quantitative variable whose
value is determined by some chance mechanism. Examples are
the total number of heads in 10 tosses of a fair coin, the toss
number of the first head in 10 tosses of a fair coin, the number
of red balls selected when drawing 2 balls (without replacement) from a container that holds 8
red balls and 20 green balls.
- Discrete and Continuous Random Variables
- Discrete random variables can only have a finite or
countably infinite number of values. The number of heads
in 10 tosses of a fair coin, the toss number of the first head
if a fair coin is tossed until a head appears, or the number
of green balls selected in the example given above.
- Continuous random variables can assume any of an uncountably
infinite set of values. For example, if a point is
picked at random on the interval from [0,1], there are an
uncountably infinite number of values that could be picked.
- Discrete random variables can only have a finite or
countably infinite number of values. The number of heads
in 10 tosses of a fair coin, the toss number of the first head
if a fair coin is tossed until a head appears, or the number
of green balls selected in the example given above.
- Probability Distribution of a Discrete Random Variable--Each
discrete random variable can only assume a finite or countably infinite number of values.
If a table is made associating the probability of each value with
the value, the table or association is the probability
distribution of that random variable. As an example consider
tossing a fair coin 3 times. A random variable that can be
associated with this experiment is a count of the number of heads
in the 3 tosses. Call this random variable T. T can
assume any of the values 0, 1, 2, or 3. The probabilities
associated with each of these values are P[T=0]=1/8, P[T=1]=3/8,
P[T=2]=3/8, and P[T=3]=1/8. These probabilities make up the
probability distribution of this random variable. In table
form, this probability distribution is:
t 0
1 2 3 P[T=t] 1/8 3/8 3/8 1/8
Notice that the sum of the probabilities is 1. This is true for any discrete random variable.Run the experiment of tossing a fair coin 3 times 1000 times updating after every 100 tosses. To do this link here, and when the page opens click the red die in front of number 4. Set the number of coins at 3. After running the experiment 1000 times, what can you say about the theoretical (in blue) and actual (in red) probability distributions of the number of heads?
In all examples of discrete random variables, the probabilities in the probability distribution table give the 'long-term' proportion of times that the random variable assumes each possible value.
- Example 1: Find the probability distribution of the number of
red balls selected if two balls are selected (without
replacement) from a container which has 4 balls numbered 1 through
4, with balls numbered 1 and 2 red balls and balls numbered 3
and 4 green balls.
- Example 2: Find the probability distribution of the number of
red balls selected if two balls are selected (with
replacement) from a container which has 4 balls numbered 1 through
4, with balls numbered 1 and 2 red
balls and balls numbered 3 and 4 green balls.
- Example 3: Find the probability distribution of the number of
red balls selected if two balls are selected (with
replacement) from a container with 20 balls, 10 of them red and 10
green.
- Example 4: Find the probability distribution of the number of
red balls selected if two balls are selected (without
replacement) from a container with 20 balls, 10 of them red and 10
green.
For examples 3 and 4, you can use this link to a simulation of the situation. When the page opens click the red die in front of number 4 to open the simulation. When the simulation opens, set N to 20, set R to 10, the number of red balls, and set n, the sample size to 2. Select with or without replacement as appropriate for the example. The blue graph and the text below it will show probabilities for each number of red balls.
- Example 5: In examples 3 and 4 decrease the number of red balls.
What happens to the probability distribution of the number of red
balls in the sample? Is this expected? Also, for each
number of red balls, observe the differences in the probability
distribution with and without replacement. Next, set the
number of red balls at 10, use sampling with or without
replacement, and run the simulation of drawing 2 balls, 1000
times, updating every 100 times. What do you see?
- Mean
(also called Expected Value) and Standard Deviation of a Discrete Random Variable
Consider again the count of heads in 3 tosses of a fair coin. If this experiment is repeated, say 10 times, and the
number of heads in each series of 3 tosses is counted, you will have a set of numbers like 0,1,3,1,2,2,1,1,3,0. The average of these numbers is the average value for the random variable, the number of heads in 3 tosses of a fair coin. To see what happens in a larger number of runs of the experiment, again link here, and when the page opens click the red die in front of number 4. Set the number of coins at 3 and run the experiment 100 times. What is the average number of heads per 3 tosses? Now reset and run the experiment 1000 times. What is the average number of heads per 3 tosses?The long-term average number of heads is called the expected value of the random variable, the number of heads in 3 tosses of a fair coin. This expected value can be found for most random variables. Think of expected value as the average value of a random variable.
There is an easier way to find the expected value of this (or any) discrete random variable. If the experiment of tossing the coin 3 times is repeated for a large number, N, times, the experiment will end in 0 heads n0 times, in 1 head n1 times, in 2 heads n2 times, and in 3 heads n3 times. The total number of heads is 0 n0 + 1 n1 + 2 n2 + 3 n3, and the average number of heads per run of the experiment is
(0 n0 + 1 n1 + 2 n2 + 3 n3)/N = 0 (n0/N) + 1 (n1/N) + 2 (n2/N) + 3 (n3/N)
For large N, (n0/N) ~ P[0 Heads], (n1/N) ~ P[1 Head], (n2/N) ~ P[2 Heads], (n3/N) ~ P[3 Heads], so the average number of heads per run of the experiment is
0 P[0 Heads] + 1P[1 Head] + 2 P[2 Heads] + 3 P[3 Heads]
This is called the Expected Value or Mean and is denoted, for a general random variable X, by E[X]. It can be computed by
Using this formula on the random variable T, the total number of heads in 3 tosses of a fair coin, you get
µ=E[T] = 0 (1/8) + 1 (3/8) + 2 (3/8) + 3 (1/8) = 12/8 = 3/2 = 1.5. This can be interpreted as the average number of heads per sequence of 3 tosses if the experiment is repeated a large number of times.
Just as you are able to find the average value for a random variable, so you can also find the standard deviation of the random variable. In the case of a random variable, the standard deviation is given by
For random variable T, the total number of heads in 3 tosses of a fair coin, the standard deviation computed by the rightmost formula is
SD[T] = Square Root of [02 (1/8) + 12 (3/8) + 22 (3/8) + 32 (1/8) - (3/2)2] = Square Root of [3/4] = 31/2/2
The Binomial Random Variable
- Definition--A
binomial random variable with parameters n and p is a count of
the number of successes in n experiments (or trials).
- Each
trial can result in only two outcomes, a success, S, or
a failure, F.
- The
probability of a success on any trial is P[S] = p and
the probability of a failure on any trial is P[F] = 1-p
= q
- The
outcome of any trial has no effect on outcomes of other
trials.
- The
binomial random variable is a count of the number of
successes in n trials.
An example is the toss of a fair coin 3 times. If you think of a success as a head, the count of the number of heads in 3 tosses satisfies the definition of a binomial random variable. Here n=3 and p=1/2.
Another example is the drawing of 2 balls with replacement from a container with 20 balls, 10 of which are red and 10 white. A red ball is considered a success. Then the count of red balls is a binomial random variable with n=2 and p=1/2. If the drawing is without replacement, the probabilities of a red ball change from draw to draw, so the count of red balls is no longer a binomial random variable. However, when the number of balls drawn is much smaller than the total number of balls in the container, the count of red balls will be distributed almost like a binomial random variable.
- Each
trial can result in only two outcomes, a success, S, or
a failure, F.
- Probability
Distribution of a Binomial Random Variable with parameters n
and p.
Since the binomial random variable is a count of the number of successes in n trials, the number of successes can only be an integer between 0 and n. Thus, if X is the total number of successes in n trials, P[X=k] will only be nonzero for k=0,1,2,3,...,n. The formula for P[X=k] is
where
- Mean and Standard Deviation for a binomial random variable with
parameters n and p.
Formulas for the mean or expected value and standard deviation of any random variable have been given above. When these formulas are applied to a binomial random variable with parameters n and p, they simplify to
the following expressions:
and 
- Examples
- According to information collected in the last census, 20%
of working women have never been married. If 10 working
women are selected at random, what is the probability that
exactly 3 of them have never been married? What is the
probability that at most 2 of them have never been married?
What is the probability that all 10 of them have never been
married? How many would you expect to be never married
women?
- If a person makes random guesses at each of the questions on
a 20 question true-false test, what is the probability that they will get exactly 10
questions correct? What is the probability that they
will get none of the questions correct? What is the
probability that they will get at least 1 correct answer?
How many questions would you expect them to get correct?
- You draw 5 balls (with replacement) from a container in
which 30 of the balls are red and 20 are green. What is the probability that none of the
balls that you pick are green? What is the probability
that at least one of the balls selected is green? What
are the mean and standard deviation of the number of red balls
selected?
- You play a game in which you toss a pair of fair dice and win a prize of $5 if the sum of numbers on the dice is 6. You play the game 12 times. What is the probability that you win $5 in exactly 3 of those games? In no game? In at least one game? How many times would you expect to win?
- According to information collected in the last census, 20%
of working women have never been married. If 10 working
women are selected at random, what is the probability that
exactly 3 of them have never been married? What is the
probability that at most 2 of them have never been married?
What is the probability that all 10 of them have never been
married? How many would you expect to be never married
women?