Download P(A and B)

Probability Rules
In the following sections, we will transition from looking
at the probability of one event to the probability of
multiple events (compound events). We will look at
two different types of compound events:
1.P(A or B)
2.P(A and B)
P(A or B)
P(A or B) denotes the probability that event A or event B or both occur in a procedure or
trial. To be an outcome of the compound event A or B, the outcome can be in either A or
B or both. If an outcome is part of both, we only count it once.
Example: Suppose that you draw a card from a standard deck of cards. Let event A be
the event that you draw a heart card and event B the event the that you draw a face card.
Use the classical approach to find each of the following probabilities.
13
12
22
a) P(A) 
b) P(B) 
c) P(A or B) 
52
52
52
Event B
S
Event A
Addition Rule
If we look back at the previous example, notice that in order to find P(A
or B) we counted all of the outcomes of A and all of the outcomes of B
and subtracted the outcomes that they had in common. This intuitive
idea can formalize with the following probability rule:
P(A or B)= P(A) + P(B) – P(A and B)
P(H or F)= P(H) + P(F) – P(H and F)
where P(A and B) denotes the probability that event A and event B
happen at the same time.
So, we could have solved our card example using the following
computation:
P(A)
+ P(B)
- P(A and B) = P(A or B)
13
12
3


52
52
52
22

52
Example Using Addition Rule
Suppose you roll two dice. Let A be the event of rolling the same number on
each die, B be the event of rolling two odd numbers and C the event that
the sum of the two dice is less than 6.
a)
What is the sample space?
Event A
Event B
Event C
b)
a) P(A)=
6
36
b) P(B)=
9
36
c)P(A or B)=
6
9
3
12



36 36 36 36
6 10
2
14
9 10
3
16






e) P(A or C)= 36 36 36 36 f) P(B or C)= 36 36 36 36
d) P(C)=
10
36
The Addition Rule and Disjoint Events(Mutually Exclusive)
Two events that do not have any common outcomes or that can not occur at
the same time are called disjoint events(mutually exclusive).
Example: Suppose you roll a die. Let event A be the event that you roll an
even number and event B the event that you roll an odd number. Are these
events disjoint?
S
A
B
2 4 6
1 3 5
Now, lets consider P(A or B).
3 3 0 3 3
P(A or B)= P(A) + P(B) – P(A and B)=      1
6
6
6
6
6
So, for disjoint events only the addition rule simplifies to
P(A or B) = P(A) + P(B)
Two Special Disjoint Sets
Recall that the complement of event A contains all of the outcomes when event A does
not occur. Hence, A and A are disjointed sets and the P(A or A )= P(A) + P(A ). Since
the union between A and make up the entire sample space the following properties are
true,
1. P(A) + P(A ) = 1
2. P(A ) = 1 - P(A)
3. P(A) = 1 – P(A )
Properties 2 and 3 allow us to find the probabilities of event A or using one or the other
respectively. In other words, if I know P(A), I do not have to worry about counting the
outcomes of S or . I can just find P( ) using property number 2.
Example: FBI data show that 62.4% of murders are cleared by arrests. We can express
the probability of a murder being cleared by an arrest as P(cleared)= 0.624. For a
randomly selected murder, fin P( cleared ).
P( cleared )= 1 – P(cleared)
= 1 - 0.624
= 0.376