Independence and
Law of Total
Probability
Lecture 7
Independence and Law of Total
Probability
Text: A Course in Probability by Weiss 4.3 ∼ 4.4
Quiz 1
Independence
Law of Total Probability
STAT 225 Introduction to Probability Models
February 2, 2014
Whitney Huang
Purdue University
7.1
Agenda
1
Quiz 1
Independence and
Law of Total
Probability
Quiz 1
Independence
Law of Total Probability
2
Independence
3
Law of Total Probability
7.2
Quiz 1, Problem 1
Independence and
Law of Total
Probability
Students at Brobbins University were asked to state whether
they like chocolate ice cream (yes or no) and whether they like
vanilla ice cream (yes or no). The partial results of the poll are
below.
32% like neither
Quiz 1
18% like vanilla only
Independence
15% like both chocolate and vanilla
Law of Total Probability
1
Draw a Venn diagram describing the ice cream flavor
preferences at Brobbins University? (5 points)
2
What percent of the students stated that they did not like
vanilla at all? (2 points)
3
What percent of those students liking chocolate, did not
also like vanilla ? (2 points)
7.3
Quiz 1, Problem 1 solution
Independence and
Law of Total
Probability
Solution.
Quiz 1
Independence
Law of Total Probability
7.4
Quiz 1, Problem 1 solution
Independence and
Law of Total
Probability
Solution.
Quiz 1
Independence
Law of Total Probability
1
7.4
Quiz 1, Problem 1 solution
Independence and
Law of Total
Probability
Solution.
Quiz 1
Independence
Law of Total Probability
1
2
By complement rule, we have 1 − .18 − .15 = .67 = 67%
7.4
Quiz 1, Problem 1 solution
Independence and
Law of Total
Probability
Solution.
Quiz 1
Independence
Law of Total Probability
1
2
By complement rule, we have 1 − .18 − .15 = .67 = 67%
3
Conditional probability!
c
∩C)
P(V c |C) = P(V
= .35
P(C)
.50 = 0.70 = 70%
7.4
Quiz 1, Problem 2
Independence and
Law of Total
Probability
In a reality television show race there are 12 participants. In
one leg of the race the top 3 finishers are immune from
elimination and will move on to the next leg. The remaining
participants will have to undergo further challenges to be able
to move on to the next leg.
Quiz 1
Independence
Law of Total Probability
7.5
Quiz 1, Problem 2
Independence and
Law of Total
Probability
In a reality television show race there are 12 participants. In
one leg of the race the top 3 finishers are immune from
elimination and will move on to the next leg. The remaining
participants will have to undergo further challenges to be able
to move on to the next leg.
Quiz 1
1
How many ways can three contestants move to the next
round without having to complete further challenges? (3
points)
Independence
Law of Total Probability
7.5
Quiz 1, Problem 2
Independence and
Law of Total
Probability
In a reality television show race there are 12 participants. In
one leg of the race the top 3 finishers are immune from
elimination and will move on to the next leg. The remaining
participants will have to undergo further challenges to be able
to move on to the next leg.
Quiz 1
1
2
How many ways can three contestants move to the next
round without having to complete further challenges? (3
points)
Independence
Law of Total Probability
If the first place finisher receives $10000, the second
place finisher receives $5000 and third place gets $2500,
how many ways can the prize money be awarded to the
contestants? (3 points)
7.5
Independence and
Law of Total
Probability
Quiz 1, Problem 2 solution
Solution.
1
2
12
3
12!
12×11×10
9!×3! = 3×2×1
12!
12 P3 = 9! = 12 × 11 ×
=
=
1320
6
= 220
10 = 1320
Quiz 1
Independence
Law of Total Probability
7.6
Motivating Example
Independence and
Law of Total
Probability
Example
You toss a fair coin and it comes up “Heads" three times. What
is the chance that the next toss will also be a “Head"?
Solution.
Quiz 1
Independence
Law of Total Probability
The chance is simply 12 just like ANY toss of the coin ⇒ What it
did in the past will not affect the current toss!
7.7
Independence and
Law of Total
Probability
Independence
Independent events
Let A and B be events of a sample space with
P(A) > 0, P(B) > 0. We say that event B is independent of
event A if the occurrence of event A does not affect the
probability that event B occurs.
Quiz 1
Independence
P(B|A) = P(B)
Law of Total Probability
If we know A and B are independent then the general
multiplication rule will reduce to a special form
The special multiplication rule
Two events A and B, are said to be independent events if
P(A ∩ B) = P(A) × P(B)
7.8
Example 20
Independence and
Law of Total
Probability
The independence of the events A and B implies that the
following are independent:
1
Ac and B
2
A and B c
3
Ac and B c
Quiz 1
Independence
Law of Total Probability
7.9
Independence and
Law of Total
Probability
Example 20
Solution.
Given A and B are interdependent
⇒ P(A|B) = P(A) , P(B|A) = P(B)
1
Quiz 1
P(Ac ∩ B) = P(Ac |B) × P(B) = [1 − P(A|B)] × P(B)
Independence
Law of Total Probability
= [1 − P(A)] × P(B) = P(Ac ) × P(B)
2
P(B c ∩ A) = P(B c |A) × P(A) = [1 − P(B|A)] × P(A)
= [1 − P(B)] × P(A) = P(B c ) × P(A)
3
P(Ac ∩ B c ) = P(Ac |B c ) × P(B c ) = [1 − P(A|B c )] × P(B c )
= [1 − P(A)] × P(B c ) = P(Ac ) × P(B c )
7.10
Independence for more than two events
Independence and
Law of Total
Probability
Consider more than two events A1 , A2 , · · · , Ak , we have two
types of independence:
Pairwise independent
P(Ai ∩ Aj ) = P(Ai ) × P(Aj ) i, j ∈ 1, 2, · · · , k i 6= j
Quiz 1
Independence
Law of Total Probability
Mutually independent
P(Ak1 ∩ Ak2 ∩ · · · ∩ Akn ) = P(Ak1 ) × P(Ak2 ) × · · · × P(Akn )
ki , kj ∈ 1, 2, · · · , k ki 6= kj 1 ≤ n ≤ k
7.11
Example 21
Chris and his roommates each have a car. Julia’s Mercedes
SLK works with probability .98, Alex’s Mercielago Diablo works
with probability .91, and Chris’ 1987 GMC Jimmy works with
probability .24. Assume all cars work independently of on
another. What is the probability that at least 1 car works?
Independence and
Law of Total
Probability
Quiz 1
Independence
Law of Total Probability
7.12
Example 21: Julia’s SLK
Independence and
Law of Total
Probability
Quiz 1
Independence
Law of Total Probability
7.13
Example 21: Alex’s Diablo
Independence and
Law of Total
Probability
Quiz 1
Independence
Law of Total Probability
7.14
Example 21: Chris’ 1987 Jimmy
Independence and
Law of Total
Probability
Quiz 1
Independence
Law of Total Probability
7.15
Example 21
Independence and
Law of Total
Probability
Solution.
Let S denotes Julia’s SLK works, M denotes Alex’s Mercielago
Diablo works, and J denotes Chris’ 1987 GMC Jimmy works
P(S ∪ M ∪ J) = 1 − P(S c ∩ M c ∩ J c ) =
1 − P(S c ) × P(M c ) × P(J c ) = 1 − .02 × .09 × .76 = .9986
Quiz 1
Independence
Law of Total Probability
7.16
Independence and
Law of Total
Probability
Law of total probability
Recall the law of partitions
Let A1 , A2 , · · · , Ak form a partition of Ω. Then, for all events B,
P(B) =
k
X
P(Ai ∩ B)
i=1
Quiz 1
Independence
Law of Total Probability
By applying the general multiplication rule, we have the law of
total probability
P(B) =
k
X
P(B|Ai ) × P(Ai )
i=1
7.17
Example 22
Acme Consumer Goods sells three brands of computers: Mac,
Dell, and HP. 30% of the machines they sell are Mac, 50% are
Dell, and 20% are HP. Based on past experience Acme
executives know that the purchasers of Mac machines will
need service repairs with probability .2, Dell machines with
probability .15, and HP machines with probability .25. Find the
probability a customer will need service repairs on the
computer they purchased from Acme.
Independence and
Law of Total
Probability
Quiz 1
Independence
Law of Total Probability
Solution.
Apply the law of total probability. Let R denotes a customer will
need service repairs on the computer, D denotes Dell, M
denotes Mac, and H denotes HP.
P(R) = P(R|D) × P(D) + P(R|M) × P(M) + P(R|H) × P(H)
= .5 × .15 + .3 × .2 + .2 × .25 = .185
7.18
Example 23
Let us assume that a specific disease is only present in 5 out of
every 1,000 people. Suppose that the test for the disease is
accurate 99% of the time a person has the disease and 95% of
the time that a person lacks the disease. Find the probability
that a random person will test positive for this disease.
Independence and
Law of Total
Probability
Quiz 1
Independence
Solution.
Law of Total Probability
Let D denotes disease, and + denote test positive.
Given: P(D) = .005 ⇒ P(D c ) = .995,
P(+|D) =
.99,
P(−|D c ) = .95 ⇒ P(+|D c ) = .05
P(+) = P(+|D) × P(D) + P(+|D c ) × P(D c ) =
.99 × .005 + .05 × .995 = .0547
7.19
Summary
Independence and
Law of Total
Probability
In this lecture, we learned
The concept of Independence
Special multiplication rule and law of total probability
Quiz 1
Independence
Law of Total Probability
7.20