Stat 330 (Spring 2015): slide set 6
E2 = choose Box 2 and pick a gold coin
E1 = choose Box 1 and pick a gold coin
Define new events E1 and E2:
P (Select a gold coin|B2) = 0.5. Why?
P (Select a gold coin|B1) = 1. Why?
2
Choosing one box (at random) means, that all boxes are equally likely to
be chosen: P (Bi) = 13 for i = 1, 2, 3.
A tree diagram shows all possible outcomes of this two-step procedure:
Treasure Hunt... continued
Last update: January 13, 2015
Stat 330 (Spring 2015)
Slide set 6
Stat 330 (Spring 2015): slide set 6
1
2
· 13 = 16 .
1
3
+ 16 = 0.5.
3
We just used the Law of Total Probability to compute the probability of
choosing a gold coin.
P ( gold coin ) =
Thus we have:
This is because those are the only ways to get a gold coin, as we’ve seen in
the tree diagram and they are mutually exclusive (disjoint)!
The probability to choose a gold coin is the sum of P (E1) and P (E2)
= P ( pick a gold coin |B2) · P (B2) =
P (E2) = P ( choose Box 2 and pick a gold coin )
= P ( pick a gold coin |B1) · P (B1) = 1 · 13 = 13 .
P (E1) = P ( choose Box 1 and pick a gold coin)
Treasure Hunt... continued
We use the definition of conditional probability to get P (E1) and P (E2)!
Stat 330 (Spring 2015): slide set 6
1
Define B1, B2, B3 to be the events that Box 1, 2 or 3 is selected randomly.
For a problem like this, that consists of a step-wise procedure, it is often
useful to draw a tree (a flow chart) of the choices we can make in each
step.
• Suppose that you select one of the boxes randomly and then select
one of the coins from this box. What is the probability that the coin you
select is a gold coin?
• Box 3 has two silver coins.
• Box 2 has one gold coin and one silver.
• Box 1 has two gold coins
Experiment: Treasure Hunt
Total Probability Law and Bayes’ Rule
Stat 330 (Spring 2015): slide set 6
P (A|Bi)P (Bi).
In our Treasure Hunt example, the events B1, B2, and B3 form a cover Ω,
as a coin can be drawn only from one of the boxes.
6
Note that event A does not intersect the event B3 as there are no gold
coins in Box 3.
P (A) = P (A|B1)P (B1) + P (A|B2)P (B2) = 1 · 13 + 12 · 13 = 0.5 as before.
=
P (Bj |A) =
Proof of Bayes’ Rule:
P (A|Bj )P (Bj )
.
k
j=1 P (A|Bj )P (Bj )
P (Bj ∩ A) P (A|Bj )P (Bj )
=
P (A)
P (A)
7
Theorem: Bayes’ Rule. If B1, . . . , Bk is a cover or partition of Ω, and A is
an event, then
P (A|Bj )P (Bj )
P (Bj |A) = k
.
j=1 P (A|Bj )P (Bj )
Stat 330 (Spring 2015): slide set 6
Bayes’ rule.
Because B1, . . . , Bk partition Ω, the events A∩B1, . . . A∩Bk are disjoint,
and ∪ki=1Ai = A where Ai = A ∩ Bi.
k
By Axiom (iii) (slide set 2 p.5), P (A) =
i=1 P (A ∩ Bi ) =
k
i=1 P (A|Bi )P (Bi ).
•
•
By definition of conditional probability P (A|Bi)P (Bi) = P (A ∩ Bi)
•
Stat 330 (Spring 2015): slide set 6
The probability
kof event A is put together as sum of the probabilities of the
intersections i=1 P (A ∩ Bi).
Defining event A = drawing a gold coin, we have
i=1
k
Proof of the Law of Total Probability:
P (A) =
Theorem: Law of Total Probability. If the collection of events B1, . . . , Bk
is a cover of Ω, and A is an event, then
5
Law of Total Probability (continued...)
We can depict event A using a Venn diagram:
Stat 330 (Spring 2015): slide set 6
Law of Total Probability (continued...)
4
• If we represent a multi-step procedure with a tree diagram, then the
branches of the tree form a cover.
• We can represent a cover using a Venn diagram:
(i) the events are mutually exclusive (i.e., Bi ∩ Bj = ∅ for i = j), and
k
(ii) the union of the events is Ω (i.e., i=1 Bi = Ω).
Definition. A collection of events B1, . . . Bk is called a cover or partition of
Ω if
Law of Total Probability
Stat 330 (Spring 2015): slide set 6
:=Cd
P (chip is defective
:=Td
it’s defective)
tester says = P (Cd|Td)
|
We will apply Bayes’ Rule, using Cd, and C̄d as cover.
8
If the test device says the chip is defective, what is the probability that the
chip actually is defective?
P ( “tester says chip is defective” | “chip is defective” ) = 0.94
P ( “tester says chip is good” | “chip is good” ) = 0.95
A given lot of chips contains 2% defective chips. Each chip is tested before
delivery. However, the tester is not wholly reliable:
Example 1.7.3: (Hofmann notes)
=
=
=
P (Cd|Td) =
9
Stat 330 (Spring 2015): slide set 6
P (Td|Cd)P (Cd)
P (Td|Cd)P (Cd) + P (Td|C¯d)P (C¯d)
0.94 · 0.02
0.94 · 0.02 + (1 − P (T̄d|C̄d)) · 0.98
0.94 · 0.02
= 0.28
0.94 · 0.02 + (1 − 0.95) · 0.98
P (Td|Cd)P (Cd)
P (Td)
Continue Example 1.7.3