Law of Total Probability
Partition of S is collection of disjoint events
{A1, A2, . . . , An} with ∪iAi = S.. Think of a
partition as a collection of all possibilities of
some phenomenon. Let B be any other event.
Then
P (B) =
X
P (B|Ai)P (Ai)
i
Bayes’ Theorem
P (B|Aj )P (Aj )
P (Aj |B) = P
i P (B|Ai )P (Ai )
1
Example 2.29 p76. Store sells three brands DVD
players. Sales: Brand 1 has 50%, Brand 2 has
30%, Brand 3 has 20%.
Warranties: Brand 1 has 25% returns, Brand
2 has 20% returns, Brand 3 has 10% returns.
Customer is selected at random.
1. What is probability that customer bought
Brand 1 and item will be returned under
warranty?
2. What is probability that an item will be
returned under warranty?
3. If an item is returned, what is conditional
probability that item is Brand 1? Brand 2?
Brand 3?
2
Ai = {customer purchased Brand i = 1, 2, 3}.
B = {DVD returned}.
..
3
Graphical representation of Bayes’ Theorem. Compute new probability relative to event B. (relative area or probability of Aj relative to B.
..
4
Example. 5% of components produced are defective (event A). Screening test indicates a
defect (event B). Test detects defectives with
probability 0.90, ie P (B|A) = 0.90 but may
also indicate not-defective components as defective (false positive) with probability P (B|A0) =
0.01. Suppose items are tested before being
shipped and only those that pass the screening test are shipped.
What percent of shipped items are defective?
P (A ∩ B 0)
0
P (A|B ) =
P (B 0)
P (B 0|A)P (A)
=
P (A ∩ B 0) + P (A0 ∩ B 0)
P (B 0|A)P (A)
=
P (B 0|A)P (A) + P (B 0|A0)P (B 0)
(0.10)(0.05)
=
(0.10)(0.05) + (0.99)(0.95)
= 0.0053
5
What percent of the indicated items are actually defective?
P (A|B) =?
Easiest to view Bayes Theorem in a table with
numbers rather than probabilities. Suppose
100,000 items produced.
A
A0
totals
B
4,500
950
5,450
B0
500
94,050
94,550
totals
5,000
95,000
100,000
P (A|B 0) = 500/94550 = 0.0053
P (A|B) = 4500/5450 = 0.8257
6
Example. Space vehicle designed to land on
Mars. Assume ground conditions on Mars are
either hard of soft. If ground hard, then probability of successful landing is 0.9, but if ground
soft, then probability falls to 0.5. Best estimates are hard ground three times more likely
than soft (i.e. odds of hard to soft are 3:1).
1. Show probability of successful landing = 0.8.
2. Suppose stick projected first to test ground.
Penetrates hard ground with probability 0.2,
penetrates soft ground with probability 0.9.
Show that the conditional probability that
ground is hard, given that stick penetrates = 0.4.
3. Show that probability of successful landing
if stick penetrates ground =0.66.
4. What is probability of a successful landing
if stick does not penetrate ground?
7
Independence
Suppose
P (A|B) = P (A)
B provides no information about P (A).
Equivalent to
P (A ∩ B) = P (B)P (A)
which imples
P (B|A) = P (B)
Call A, B independent.
Equivalent: A0, B, or A0, B 0, or A, B 0 independent.
Events A1, . . . , An are independent if
P (∩n
i=1 Ai ) =
n
Y
P (Ai)
i=1
8
Example. 30% of company’s washing machines
require service under warranty, 10% of its dryers need such service. If someone purchases
both a washer and a dryer made by this company, what is the probability that both machines will need warranty service?
A = {washer needs service}. P (A) = 0.3.
B = {dryer needs service}. P (B) = 0.1.
P (A ∩ B) = P (A)P (B) = 0.03.
Example. Given P (A1) = 0.95, P (A2) = 0.95,
A1, A2 independent, find P (A1 ∪ A2).
P (A1 ∪ A2) = P (A1) + P (A2) − P (A1 ∩ A2)
= 0.95 + 0.95 − (0.95)(0.95)
= 0.9975
9
Reliability. Reliability of a device or system =
probability operates for a specified duration.
Parallel circuit provides redundancy and increases
reliability.
Parallel circuit.
fig_02_21
Reliability = P (A1 ∪ A2)
= 1 − P (A1 ∩ A2)
= 1 − (0.05)(0.05)
= 0.9975
10
Series circuit. Sometimes more than one
device must operate for a system to
operate.
Reliability of two devices given by P (A1) =
0.80, P (A2) = 0.90. Operate independently. System reliability = P (A1 ∩ A2)
Reliability = P (A1 ∩ A2)
= (0.80)(0.90) = 0.72
fig_02_20
11
Mixed circuits. Decompose into smaller series,
parallel circuits.
Example 2.36 p85. Article “Reliability Evaluation of Solar Photovoltaic Array” (Solar Energy, 2002: 129-141) presents various configurations of solar photovoltaic arrays consisting
of crystalline silicon solar cells.
Configuration (a).
12
Reliability of upper circuit = P (U )
= P (A1 ∩ A2 ∩ A3)
= P (A1)P (A2)P (A3) = 0.93 = 0.729
Reliability of lower circuit = P (L)
= P (A4 ∩ A5 ∩ A6)
= P (A4)P (A5)P (A6) = 0.93 = 0.729
Reliability of circuit = P (L ∪ U )
= 1 − P (L0 ∩ U 0)
= 1 − (1 − 0.729)(1 − 0.729)
= 0.9266
Alternate calculation.
P (L ∪ U ) = P (L) + P (U ) − P (L ∩ U )
= 0.729 + 0.729 − (0.729)(0.729) = 0.9266
13
Configuration (b). Connecting ties across each
column of junctions. Now the system fails as
soon as an entire column fails, and system
works only if both columns work.
P (1 or 4 work) = P (A1 ∪ A4)
= 1 − (1 − .99)(1 − .99) = 0.9900
P (2 or 5 work) = 0.9900
P (3 or 6 work) = 0.9900
Reliability of circuit = 0.99003 = 0.9703
14
Practice Circuits. In the next three diagrams,
the numbers represent the respective reliabilities (probability of working) of the individual
components. Find the reliability of each circuit.
fig_02_23
15
fig_02_24
16
fig_02_22
17
In the next two diagrams, the numbers represent the failure probabilities of the individual
components. Find the reliability of each circuit.
fig_02_25
18
fig_02_26
19