Probability of an Event
Gregorio P. Agatep, Jr.
Mathematics Department
INNHS
Probability of an Event
Illustrations:
An experiment is a procedure that leads a)Tossing a coin; b) throwing a die
to well-defined results called outcomes. c) Drawing a letter from the English
alphabet
b)
a)
The set of all outcomes is called the
c)
sample space denoted by S.
Every element of the sample space is
called a sample point
A subset of S containing one or more
elements of the sample space is called
an event.
Probability is the chance that an event
occurs. It can be expressed as a
fraction, decimal, or a percentage.
A = event of getting a tail in 1 toss of
a coin
A = {T}
B = event of getting a prime number
is 1 throw of a die
B = {2, 3, 5}
C = event of drawing a vowel
C = {a, e, i, o, u}
Notations:
Let A and B be events.
P(A) = probability of event A
P(S) = probability of the sample space
AB = intersection of events A and B
P(AB) = probability of the intersection of
events A and B
AB = union of events A and B
P(AB) = probability of the union of events A
and B
A’ = complement of event A
P(A’) = probability of the complement of A.
If the sample space for an
experiment has elements, all are equally
likely to occur, each element in the sample
space has a probability equal to . If A, with
elements, is any event in this sample
space, then the probability of event A is
P(A) =
(Formula 1)
Remark:
Formula 1 is called Classical Probability
Example 1: A box contains 4 cloud 9, 5 choco mucho, and 3 beng-beng chocolates. A
chocolate is taken out from the jar at random, what is the probability that a
a. cloud 9 is taken?
b. beng-beng or choco mucho is taken?
Solution:
Let A = event that a cloud 9 is taken
B = event that a beng-beng is taken
C = event that a choco mucho is taken
a) Using Formula 1,
4
1
P(A) =
=
12
and
𝑁 = 12
3
b) Using Formula 1, and
P(BC)=
𝑁 = 12
Example 2: A coin bank contains 5 ten-peso coins, 9 five-peso coins, and 11 one-peso coins. What is the
probability of getting a
a) ten-peso coin out of the coin bank?
b) five-peso coin out of the coin bank?
c) a one-peso coin or a ten-peso coin out of the coin bank?
d) a five-peso coin or a one-peso coin out of the coin bank?
Solution:
Let A = event that a ten-peso coin is taken out of the coin bank
B = event that a five-peso coin is taken out of the coin bank
C = event that a one-peso coin is taken out of the coin bank
a) Using Formula 1, and
P(A) =
b) Using Formula 1, and
Remark: To be successful in solving probability problems,
1) define the events properly; and
P(B) =
2) symbolize correctly the conditions (given and
c) Using Formula 1, and
unknown).
P(AC) =
d) Using Formula 1, and
P(BC) =
Properties of Probability
Let S be the sample space and A is
an event in S. Then
1.
0 P(A) 1
2.
P() = 0
3.
P(S) = 1
Example 3: A die is biased in such a way that
an odd number is thrice as likely to occur as an
even number. If A is the event that a number
more than 2 occurs in a single toss of a die, find
P(A).
Solution:
The sample space S = {1, 2, 3, 4, 5, 6}.
Assign as the probability of an even number
and as the probability of an odd number.
Since P(S) = 1, we have
P(S) = = 1
=1
A = {3, 4, 5, 6}
P(A) =
=
Example 4: A die is biased in such a way that an odd number is thrice as likely to
occur as an even number. Let A be the event that a prime number turns up and B be
the event that a number less than 4 occurs. Find : (a) P(AB); (b) P(AB)
Solution:
The sample space S = {1, 2, 3, 4, 5, 6}. Assign as the probability of an even
number and as the probability of an odd number. Since P(S) = 1, we have
P(S) = = 1
=1
Now,
A = {2, 3, 5}
B = {1, 2, 3}
a) AB =
{2, 3}.
P(AB) =
b) AB =
{1, 2,
P(AB)
= 3, 5}.
Example 5: In a poker hand of 5 cards, find the probability of holding 3 queens and
2 kings.
Solution:
The number of ways of having 3 queens from 4 is
and the number of ways of having 2 kings from 4 is
.
Using the Fundamental Principle of Counting, there are .
So, there are 24 hands with 3 queens and 2 kings.
Next, the total number of 5-card poker hand is .
Thus, the probability of event A of getting a hand with 3 queens and 2 kings is
P(A) =