Objectives (BPS chapter 12)
General rules of probability

Independence and the multiplication rule

The general addition rule

Conditional probability

The general multiplication rule

Independence

Tree diagrams
Independent versus disjoint events
Two events are independent if the probability that one event occurs
on any given trial of an experiment is not influenced in any way by
the occurrence of the other event.
Imagine coins spread out so that half were heads up, and half were tails up.
Pick a coin at random. The probability that it is heads-up is 0.5. But, if you don’t
put it back, the probability of picking up another heads-up coin is now less than
0.5. Without replacement, successive trials are not independent.
In this example, the trials are independent only when you put
the coin back (“sampling with replacement”) each time.
The multiplication rule for independent events
Two events A and B are independent if knowing
that one occurs does not change the probability
that the other occurs. If A and B are
independent, P(A and B) = P(A) P(B)
If A is the event that you roll a “1” on a red 6-sided die, and B is the event that you
roll a “1” on a blue 4-sided die, what is the probability that if you roll both at the
same time, you will roll a “1” on both dice? Since the two dice are independent,
P(both “1”s) = P(“1” on red die) P(“1” on blue die)
= (1/6) (1/4) = 1/24
The general addition rule
General addition rule for any two events A and B:
The probability that A occurs,
or B occurs, or both events occur is:
P(A or B) = P(A) + P(B) – P(A and B)
What is the probability of randomly drawing either an ace or a heart from a pack of
52 playing cards? There are 4 aces in the pack and 13 hearts. However, one card
is both an ace and a heart. Thus:
P(ace or heart) = P(ace) + P(heart) – P(ace and heart)
= 4/52 + 13/52 - 1/52 = 16/52 ≈ 0.3
Conditional probability
Conditional probabilities reflect how the probability of an event can
change if we know that some other event has occurred/is occurring.

Example: The probability that a cloudy day will result in rain is different if
you live in Los Angeles than if you live in Seattle.

Our brains effortlessly calculate conditional probabilities, updating our
“degree of belief” with each new piece of evidence.
The conditional probability
of event B given event A is:
(provided that P(A) ≠ 0)
P( A and B)
P( B | A) 
P( A)
If A and B are independent, P(B | A) =P(B).
The general multiplication rule

The probability that any two events, A and B, both occur is:
P(A and B) = P(A)P(B|A)
This is the general multiplication rule.

If A and B are independent, then P(A and B) = P(A)P(B)
A and B are independent when they have no influence on each other’s occurrence.
Another way of showing this is P (B | A) = P (B) or P (A | B) = P (A).
What is the probability of randomly drawing either an ace or heart from a pack of
52 playing cards? There are four aces in the pack and thirteen hearts.
P(heart|ace) = 1/4
P(ace) = 4/52
P(ace and heart) = P(ace)* P(heart|ace) = (4/52)*(1/4) = 1/52
Notice that heart and ace are independent events.
P(ace or heart) =P(ace)+P(heart)-P(ace and heart)=4/52+13/52-1/52=16/52
Probability trees
Conditional probabilities can get complex and it is often a good strategy
to build a probability tree that represents all possible outcomes
graphically and assigns conditional probabilities to subsets of events.
Tree diagram for chat room
habits for three adult age
groups.
Internet
user
0.47
P(chatting) = 0.136 + 0.099 + 0.017
= 0.252
About 25% of all adult Internet users visit chat rooms.
Breast cancer screening
If a woman in her 20s gets screened for breast cancer and receives a positive
test result, what is the probability that she has breast cancer?
Diagnosis
sensitivity 0.8
Disease
incidence
0.0004
Positive
Cancer
0.2
Mammography
0.9996
0.1
Negative False negative
False positive
Positive
No cancer
Incidence of breast
cancer among
women ages 20–30
0.9
Diagnosis
specificity
Negative
Mammography
performance
She could either have a positive test and have breast cancer, or have a positive
test but not have cancer (false positive).
Diagnosis
sensitivity 0.8
Disease
incidence
Positive
Cancer
0.0004
0.2
Mammography
0.1
0.9996
Negative
False negative
Positive
False positive
No cancer
Incidence of breast
cancer among
women ages 20–30
0.9
Diagnosis
specificity
Negative
Mammography
performance
Possible outcomes given the positive diagnosis: positive test and breast cancer,
or positive test but not cancer (false positive).
P(cancer | pos) 

P(cancer and pos)
P(cancer and pos)  P(nocancer and pos)
0.0004*0.8
 0.3%
0.0004*0.8  0.9996*0.1
This value is called the positive predictive value, or PV+. It is an important piece
of information but, unfortunately, is rarely communicated to patients.
Practice Problem 1
Musical styles other than rock and pop are becoming more
popular. A survey of college students finds that 40% like country
music, 30% like gospel music, and 10% like both country and
gospel music.
A: likes country music, P(A)=0.4
B: likes gospel music, P(B)=0.3
A and B: likes both, P(A and B)=0.1

What is the probability that a randomly selected college student
likes gospel but not country music?
P( Ac  B)  0.2
Practice Problem 1

What is the probability that a randomly selected college student
doesn’t like gospel or country music?
P( A  B )  1  P( A  B)  1  0.6  0.4
c

c
What is the probability that the student chosen likes gospel music
given that he/she likes country music?
P( A  B) 0.1
P( B | A) 

 0.25
P( A)
0.4
Practice Problem 2

You and a friend decide to have lunch at a nearby Chinese restaurant. The table
below lists the six meals you and your friend ever order, along with the price of the
dish, the probabilities that you will order each dish, and the probabilities your friend
will order each dish. Assume that your choice of a meal has no bearing on your
friend’s choice (and vice versa).
Luncheon special
Price
Your
probability
Your friend’s
probability
Roast Pork with
Bean Sprouts
$4.00
0.05
0.20
Szechuen Chicken
(spicy)
$4.50
0.10
0.10
Curry Shrimp
(spicy)
$4.75
0.30
0.30
Beef with Broccoli
$4.50
0.10
0.10
Chicken Wings
$3.25
0.10
0.10
Delight of Three
$5.00
0.35
0.20
Practice Problem 2
1. Let A be the event that you order a spicy lunch. Find P(A).
1.
P(A)= P(Szechuen Chicken or Curry Shrimp) =0.1+0.3=0.4
2. Let B be the event you do not order a spicy lunch, and C be the event you order a
lunch that costs less than $4.60. Find P(B and C).
P (B and C) = P(Roast Pork with Bean Sprouts or Beef with Broccoli or Chicken
Wings)
= 0.05 + 0.1+ 0.1=0.25
3. With B and C defined in the previous part, find P(B or C).
P(B or C) = P(𝐵) + P(C) − P(B and C) = 0.6 + 0.35 − 0.25 = 0.7
4. What is the probability that both you and your friend order spicy lunches?
P(A and E) P(A)P(E) = 0.4 × 0.4 = 0.16