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Thinking
Mathematically
Fundamentals of Probability
Computing Theoretical
Probability
number of outcomes in event E
P( E ) 
number of outcomes in sample space S
Example of Theoretical
Probability
What is the probability of getting at most 2
heads when a coin is tossed 3 times?
Solution
E  {TTT , THT , HTT , TTH , HHT , HTH , THH }
n( E )  7
S  {TTT , THT , HTT , TTH , HHT , HTH , THH , HHH }
n( S )  8
n( E ) 7
P( E ) 

n( S ) 8
The probability of getting at most 2 heads
when a coin is tossed 3 times is 7/8
Example Computing Theoretical
Probability
A die is rolled once. Find the Probability of
getting a number less than 5.
Solution
The event of getting a number less than 5 can
occur in 4 ways: 1, 2, 3, 4.
P(less than 5) =
(number of ways a number less than 5 can occur)
(total number of possible outcomes)
= 4/6 = 2/3
Example Probability and a Deck of 52
Cards
You are dealt one card from a standard 52card deck. Find the probability of being
dealt a King.
Solution
Because there are 52 cards, the total number
of possible ways of being dealt a single card
is 52. We use 52, the total number of
possible outcomes, as the number in the
denominator. Because there are 4 kings in
the deck, the event of being dealt a king can
occur 4 ways.
P(king) = 4/52 = 1/13
Example Probabilities in Genetics
Each person carries two genes that are related to
the absence or presence of the disease cystic
fibrosis. 1 in 25 Americans carries one normal
gene and one defective gene. If we use c to
represent a defective gene and C a normal gene,
such a carrier can be designated as Cc. Thus, CC
is a person who neither carries nor has cystic
fibrosis, Cc is a carrier who is not actually sick,
and cc is a person sick with the disease. One copy
of each gene is passed on to the child from the
parents. If each parent carries one cystic fibrosis
gene, what is the probability that their child will
have cystic fibrosis?
Solution
The table shows the four equally likely outcomes.
Only the cc child will develop cystic fibrosis.
Thus, P(cystic fibrosis) = 1/4
If each parent carries one cystic fibrosis gene, the
probability that their child will have cystic fibrosis
is 1/4.
Second Parent
C
First
C CC
Parent c Cc
c
Cc
cc
Empirical Probability
observed number of times E occurs
P( E ) 
total number of observed occurrences
Example Computing Empirical
Probability
There are approximately 3 million Arab Americans
in America. The circle graph shows that the
majority of Arab Americans are Christians. If an
Arab American is selected at random, find the
empirical probability of selecting a Catholic.
0.36
Muslim
0.69
Catholic
0.69
Protestant
1.26
Greek
Orthodox
Solution
The probability of selecting a Catholic is the
observed number of Arab Americans who are
Catholic, 1.26 (million), divided by the total
number of Arab Americans, 3 (million).
P(selecting a Catholic from the Arab
American Population) = 1.26/3 = 0.42
Thinking
Mathematically
Fundamentals of Probability