14.7 Conditional Events
Objective: Students will be able to find the
probability of an event given the occurrence
of another event.
Suppose John has drawn a card from a standard
deck of 52 cards. What is the probability that he
picks a queen?
Now, suppose John tells you that he drew a red
card? With this new information, what is the
probability he has drawn a queen?
When you need to find the probability of an
event under the condition that some preceding
event has occurred, you are finding the
conditional probability.
The conditional probability of a queen given
that it is known to be a red card can be
symbolized as P(queen | red card)
Conditional Probability:
The conditional probability of event A, given
event B, is defined as
𝑃(𝐴|𝐵) =
𝑃(𝐴 𝑎𝑛𝑑 𝐵)
𝑃(𝐵)
where P(B) ≠ 0
So, P(queen | red card) can be found as:
2
P(queen and red) = 52
26
P(red card) = 52
P(queen | red card)
=
=
𝑃(𝑞𝑢𝑒𝑒𝑛 𝑎𝑛𝑑 𝑟𝑒𝑑)
𝑃(𝑟𝑒𝑑 𝑐𝑎𝑟𝑑)
2
52
26
52
2
52
= 52 ∙ 26
1
= 13
The set of all possible outcomes for an event is
called the sample space.
A reduced sample space is the subset of a
sample space that contains only those outcomes
that satisfy a given condition.
For example, if a pair of dice is thrown, the
sample space would contain 36 outcomes of the
form (1, 1), (1, 2), (1, 3), …, (6, 6):
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
Under the condition that the dice must match,
the reduced sample space would contain only
6 outcomes: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).
Example: A pair of dice is thrown. Find the
probability that the dice match, given that
their sum is greater than 5.
Sum greater than 5:
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
The reduced sample space is 26 outcomes.
But also if the dice match:
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
4
2
So the answer is 26 or 13.
Example: A pair of dice is rolled. Find the
probability that the dice match, given their
sum is greater than 9.
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
Sum greater than 9:
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
Both match:
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
2
1
So the probability is 6 or 3.
In some situations, event A is a subset of
event B. When this occurs, the probability
that both event A and event B occur is the
same as the probability of event A occurring.
In these situations,
𝑃(𝐴)
𝑃(𝐴|𝐵) = 𝑃(𝐵)
Example: One card is drawn from a standard
deck of 52 cards. What is the probability that
it is a queen if it is known to be a face card?
Let event A be that a queen is drawn.
Let event B be that a face card is drawn.
Since a queen IS a face card,
A is a subset of B.
Therefore P(A and B) = P(A)
4
1
12
3
P(A) = 52 = 13
P(B) = 52 = 13
𝑃(𝐴)
P(A|B) = 𝑃(𝐵)
=
1
13
3
13
1 13
= 13· 3
1
=3
HW: p787 #7 – 34