Basic Rules of Probability
To work with probability in this course we must now learn a few rules as well as a
few more definitions.
 Rule 1: The probability P(A) of any event A satisfies 0≤ P(A) ≤ 1. In other
words, all probabilities must be some number that is no less than 0 and no
more than 1. A probability of “0” means that the event will never happen
and a probability of “1” means that the event will happen every time a trial
is performed. .
 Rule 2: If S is the sample space in a probability model, then P(S) = 1.
What this is saying is that the total of all the probabilities in a distribution
can never be larger than the sample space.
 Rule 3: The complement of any event A is the event that A does not
occur, written as A'. The complement rule states that P(A')=1-P(A). Here
is an example: If the probability of passing a class is, say, 0.90 then the
probability of not passing the class is 1 – 0.90 = 0.10.
 Rule 4: Two events A and B are disjoint (aka. mutually exclusive) if they
have no outcomes in common and so can never occur simultaneously.
P(A or B) = P(A) + P(B)
Here is an example that may shed some light on how these rules can help in
solving problems:
Faked numbers in tax returns, payment records, invoices, expense account
claims, and many other settings often display patterns that aren’t present
in legitimate records. The first digits of numbers in legitimate records
often follow a distribution known as Benford’s Law. Here is the
distribution:
First Digit:
1
2
3
4
5
6
7
8
9
Probability: 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046
Use the table and the rules above to find the following probabilities:
1. P(the first digit is 1) The answer is 0.301.
2. P(the first digit is 1 or more)
Notice that the table above accounts for all digits one or greater. And
according to rule 2 above the sample space must equal 1…which
translates to the fact that the sum of all the probabilities in the table
must be 1, since the table accounts for all possibilities. So, the
answer to this question is 1
3. P(the first digit is 6 or greater)
Since there are four opportunities for the first digit to be 6 or greater
(i.e., 6, 7, 8, or 9) then we want to account for all of them. They are
also all disjoint (mutually exclusive) in that a 6 is not the same as a 7
which is not the same as an 8, etc. So, according to rule 4 above the
answer is 0.067 + 0.058 + 0.051 + 0.046 = 0.222
4. P(the first digit is 5 or less)
There are a couple of ways to look at this. Rule 3 above allows us to
use complements. The event “5 or less” is the opposite of the event
“6 or more”. So, since the sum of all the probabilities in the table
must be 1 and since we already know from problem 3 that the
probability of the first digit being 6 or more is 0.222 it should follow
that the probability of the first digit being 5 or less is 1 – 0.222 =
0.778
Another way we write out probabilities is with set notation. For example, the
event {AUB} is read as “A union B” and is the set of all outcomes that are either
in A or in B. The symbol for an intersection of sets is “∩” (an upside down “U”)
and means all the elements the sets share. Let’s look at an example of this using
the same table from above
Faked numbers in tax returns, payment records, invoices, expense account
claims, and many other settings often display patterns that aren’t present
in legitimate records. The first digits of numbers in legitimate records
often follow a distribution known as Benford’s Law. Here is the
distribution:
First Digit:
1
2
3
4
5
6
7
8
9
Probability: 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046
It has been shown that:
P(A) = P(first digit is 1) = 0.301
P(B) = P(first digit is 6 or greater) = 0.222
P(C) = P(first digit is odd) = 0.609
We will define event D to be {first digit is less than 4}. Using the union and
intersection notation, find the following probabilities.
a)
b)
c)
d)
e)
P(D)
P(B U D)
P(Dc)
P(C ∩ D)
P(B ∩ C)
Answers with explanations:
a) P(D) = P(the first digit is less than 4) = P(1) + P(2) + P(3) = 0.301 +
0.176 + 0.125 = 0.602
b) P(B U D)
This is asking for the probability of the first digit being 6 or greater or
less than 4. That means that the first digit can be either 1 or 2 or 3 or
6 or 7 or 8 or 9. Since they are all disjoint we add the probabilities
together and get 0.824
c) P(Dc)
Can you see now that this can be calculated by 1- P(D) = 1 – 0.602 =
0.398
d) P(C ∩ D)
This is asking for the probability of all the possibilities that event C
and D have in common. Event C is that the first digit is odd (1, 3, 5, 7,
or 9). Event D is that the first digit is less than 4 (1, 2, or 3). Look at
the two sets. They both have 1 and 3 in common. So, P(C ∩ D) =
0.301 + 0.125 = 0.426
e) P(B ∩ D)
This is asking for the probability of all the possibilities that event B
and C have in common. Event D is that the first digit is less than 4 (1,
2 or 3). Event B is that the first digit is 6 or greater (6, 7, 8, or 9). Look
at the two sets. They have nothing in common. So, P(B ∩ D) = 0
Notice that this is different than question “b” above
Moving forward in our probability studies, it is now time to look at the
Independence and Multiplication rules.
Two events are independent if knowing that one occurs does not change the
probability that the other occurs. For example, let’s say you toss a coin and it
lands “heads”. The probability of that happening is 0.5. Now you flip the coin
again and get heads. The probability for that outcome was also 0.5. The two
events were independent because the probability of getting a heads on the
second try had nothing to do with the probability of getting the a head on the first
try.
Rule 5: If A and B are independent, then P(A and B)=P(A)P(B).
This is the multiplication rule for independent events.
Notice that the statement starts with an “if”. That means that you have to know
that events are independent before you can use this rule.