INDENG 241 risk, data analysis and simulations
Week 1. Conditioning and independence
Thibaut Mastrolia
IEOR, UC Berkeley
Fall 2025
Conditional probability
Bayes formula
Independence
Contents
1
Conditional probability
2
Bayes formula
3
Independence
Thibaut Mastrolia
INDENG 241 risk, data analysis and simulations Week 1. Conditi
Conditional probability
Toss 2 fair dice, so that each of the 36 possible outcomes is equally
1
.
likely to occur and hence has probability 36
We observe the result of the first die which is a 3.
given this information, what is the probability that the sum of the 2
dice equals 8?
Given that the initial die is a 3, there can be at most 6 possible outcomes
of our experiment, namely,
p3, 1q, p3, 2q, p3, 3q, p3, 4q, p3, 5q, p3, 6q
and the probaiblity of getting each of them is 16 conditional to the first die
is a 3, whereas the (conditional) probability of the other 30 points in the
sample space is 0.
Let E and F denote, respectively, the event that the sum of the dice is 8
and the event that the first die is a 3, then the probability just obtained is
called the conditional probability that E occurs given that F has occurred
and is denoted by PpE |F q.
Conditional probability
Definition
Consider two events E and F such that PpF q ‰ 0. Then
PpE |F q “
PpE X F q
.
PpF q
Multiplication rule
Proposition
Let pEi q1ďiďn be a family of n events with non zero probability. Then
PpE1 X ¨ ¨ ¨ X En q “ PpE1 qPpE2 |E1 qPpE3 |E1 X E2 q . . . PpEn |E1 X ¨ ¨ ¨ X En´1 q.
Bayes formula
Proposition
Let E and F be two events. Then
PpE |F q “ PpF |E q
PpE q
PpF q
and
PpE q “ PpE X F q ` PpE X F c q
“ PpE |F qPpF q ` PpE |F c qPpF c q
“ PpE |F qPpF q ` PpE |F c qp1 ´ PpF qq.
Conditional probability
Bayes formula
Independence
Bayes formula and law of total probability
Ť
Assume that Ei “ Ω such that pEi qiě1 is a sequence of mutually exclusive events. Then for any event A we have
n
ÿ
PpAq “
PpA|Ei qPpEi q.
i“1
Thibaut Mastrolia
INDENG 241 risk, data analysis and simulations Week 1. Conditi
Example
Example 3a Chapter 3 Ross textbook
An insurance company believes that people can be divided into two
classes: those who are accident prone and those who are not. The
company’s statistics show that an accident-prone person will have
an accident at some time within a fixed 1-year period with probability 0.4, whereas this probability decreases to 0.2 for a person
who is not accident prone. If we assume that 30 percent of the
population is accident prone, what is the probability that a new
policyholder will have an accident within a year of purchasing a
policy?
Example: epidemiology
Epidemiology and test policy
Percentage of the population having the disease: 0.1%.
Diagnostic accuracy: a test enables health professional to
detect it.
True-positive results: subjects positive to the detection test
who have the disease. 99%.
False-positive results (false alarm): subjects positive to the
detection test who do not have the disease. 1%.
If a person gets a positive test, what is the probability that this
person has the disease?
1. Define good notations.
We define by D the event “the person has the disease”, D c the event “the
person has not the disease”; T the event “the test is positive” and T c the
event “the test is negative”.
2. Translate the exercise with you notations.
PpDq “ 0.1% and so PpD c q “ 99.9%.
Diagnostic accuracy
True positive: PpT |Dq “ 99%.
False-positive results (false alarm): PpT |D c q “ 1%.
Question: compute PpD|T q.
3. Solution.
PpT |DqPpDq
PpT |DqPpDq ` PpT |D c qPpD c q
99% ˆ 0.1%
“
99% ˆ 0.1% ` 1% ˆ 99.9%
« 9%.
PpD|T q “
4. Extension and interpretation.
Compute this probability by now assuming that PpDq “ 10%.
Independence
The previous examples of this chapter show that the conditional
probability of E given F denoted by PpE |F q is not generally equal to
PpE q, the unconditional probability of E.
ãÑ Knowing that F has occurred generally changes the chances of E’s
occurrence.
In very special case, PpE |F q does in fact equal PpE q, meaning that
F has no impact on the frequency that E occurs.
We say that E and F are independent if PpE |F q “ PpE q in other words
Definition
Two events E and F are said to be independent if
PpE X F q “ PpE qPpF q.
Two events E and F that are not independent are said to be dependent.
Independence
Proposition
If E and F are independent, then so are E and F c .
Thus, if E is independent of F, then the probability of E’s occurrence is
unchanged by information as to whether or not F has occurred
Example 4c
Example 4c
Suppose that we toss 2 fair dice. Let E1 denote the event
that the sum of the dice is 6 and F denote the event that
the first die equals 4. Are E1 and F independent?
Now, suppose that we let E2 be the event that the sum of
the dice equals 7. Is E2 independent of F ?