Transcript Lecture 10, January 28, 2004

Statistical Independence
•Two events are statistically independent
if the probability of any one of them is
unaffected by the occurrence of the other.
•Independence can ordinarily be assumed
from the nature of the random experiment
For instance, consider Experiment 1
involving separate tosses of two fair coins,
a dime and a quarter.
Statistical Independence
• The assumed statistical independent events:
-head of the dime (Hd)
-head for the quarter (Hq)
• For Experiment 2, (which has a statistically
dependent events), the dime is tossed first. If
head is obtained (Hd), a fair quarter is tossed.
But, if tail is obtained (Td), a crooked twoheaded quarter is tossed instead.
Statistical Independence
Pr[Hq] = ½ , when Hd occurs
Pr[Hq] = 1 when Td occurs and
Pr[Hq] = 3/4, in any repetition of the
experiment
The Multiplication Law for
Independence Event
• Just to remind you that for independents
events, one outcome is not dependent to
occurrence of the next outcome.
• The second basic law (multiplication law)
of probability is intended for computing
joint probabilities.
• Pr[A and B] = Pr[A] X Pr[B]
The Multiplication Law for
Independence Event
• To illustrate, suppose a perfectly balanced
coin id fairly tossed twice in succession.
• H1, T1, H2, T2 are the events
• Probability of two heads is the probability
of the joint event H1 and H2
• Pr[H1 and H2] = Pr[H1] X Pr[H2]
=½X½=¼
An Example
•
Among 24 invoices prepared by a billing
department, 4 contain errors while the others
do not. If we randomly check 2 of these
invoices, what are the probabilities that
(a) both will contain errors
(b) neither will contain an error
Solution
(a) The probability of getting an error on the first draw is
4/24
As we pick only 2 invoices and want to have error in both
invoices, so after the first pick (with error), we have
only 23 invoices left. And the among those 23 invoices
only 3 invoices could have errors.
So, the probability of the second event (picking an invoice
with error in second draw) is 3/23.
The probability that both will contain error is
4/24 X 3/23 = 1/46 = 0.0217
Solution
(b) The probability that neither will contain
error is:
20/24 X 19/23 = 95/138 = 0.688.
Conditional Probabilities
• Consider how the probability value of one
event is affected by the occurrence of
another
• A probability value computed under the
assumption that the another event is going
to occur is a conditional probability.
An Example
It is more likely to rain tomorrow if there is a solid cloud cover
than if there are patches of blue.
For the events “rain” and “blue patches”, the following value
might apply:
Pr[rain|blue patches] = 0.30
The event “rain” is listed first, and .30 is the conditional
probability of that event; the condition - given event – “solid
overcast” is listed second, and the vertical bar is a
shorthand representation for “given that”.
Should no stipulations be made regarding sky conditions. We
might have
Pr[rain] = 0.20 which is unconditional probability
Conditional Probability
• It may be possible to
compute a conditional
probability directly
from known
probability values
using the following:
Pr[ A | B] 
Pr[ AandB]
Pr[B ]
The General Multiplication Law
Suppose that a quality-control inspector has established
that 10% of all electric assemblies fail (F) the circuit test.
Twenty percent of the failing items have poor (P) solder
joints.
This experience provides that the following particular
assembly will fail and have poor solder joints:
Pr[F] = 0.10
Pr[P|F] = 0.20
The multiplication law gives the joint probability that a
particular assembly will fail and have solder joints
Pr[F and P] = Pr[F] X Pr[P|F] = .1 X .2 = .002
The General Multiplication Law
Pr[A and B] = Pr[A] X Pr[B|A]
And
Pr[A and B] = Pr[B] X Pr[A|B]
Notice that the probability for the first event in the
product is an unconditional one, whereas the
second is a conditional probability for the other
event, given the first.
PROBABILITY TREES
• Probability trees are useful to
– calculate probabilities
– identify all simple events
– visualize the relationship among the events
• Probability trees are useful if it is possible to
– break down simple events into stages
– identify mutually exclusive and exhaustive events at
each stage
– ascertain the probabilities of events at each stage
PROBABILITY TREES
• A probability tree consists of some nodes and branches
• Nodes
– an initial unlabelled node called origin
– other nodes, each labeled with the event represented
by the node
PROBABILITY TREES
• Branches
– each branch connect a pair of nodes.
– a branch from A to B implies that event B may occur
after event A
– each branch from
• origin to A is labeled with probability P(A)
• A to B is labeled with the probability P(B|A)
PROBABILITY TREES
• Any path through the tree from the origin to a terminal
node corresponds to one possible simple event.
• All simple events and their probabilities are shown next to
the terminal nodes.
BENDRIX COMPANY’S SITUATION
• The Bendrix Company supplies contractors with materials
for the construction of houses.
• Bendrix currently has a contract with one of its customers to
fill an order by the end of July.
• There is uncertainty about whether this deadline can be
met, due to uncertainty about whether Bendrix will receive
the materials it needs from one of its suppliers by the
middle of July. It is currently July 1.
• How can the uncertainty in this situation be assessed?
PROBABILITY TREES
• Example 1: Construct a probability tree diagram for the
Bendrix Company.
Stage 1
P(
B)
=2
/3
B
Stage 2
=3/4
P(A|B)
P(A C|B
)=1/4
C
B
P(
C )=1/5
A
BA
P(BA)=0.5000
AC
BAC
P(BAC)=0.1667
A
BCA
P(BCA)=0.0667
AC
BCAC
P(BCAC)=0.0266
P(A|B
3
1/
)=
BC
Simple Joint
Events Probabilities
P(A C|B C
)=4/5
PROBABILITY TREES
• Example 2: In a bag containing 7 red chips and 5 blue
chips you select 2 chips one after the other without
replacement. Construct a probability tree diagram for this
information.
Stage 1
Stage 2
=6/11
)
R
|
R
(
P
P(
R
)=
7/
12
R
P(B|R)
=5
/
=5
B)
P(
=7/11
P(R|B)
/11
B
Simple Joint
Events Probabilities
RR
P(RR)=7/22
B
RB
P(RB)=35/132
R
BR
P(BR)=35/132
B
BB
P(BB)=5/33
12
R
P(B|B)
=4
/11
PROBABILITY TREES
• What is the probability of getting a red chip first and then a
blue chip?
• What is the probability of getting a blue chip first and then
a red chip?
• What is the probability of getting a red and a blue chip?
• What is the probability of getting 2 red chips?
Conditional Probability
The four attendants of a gasoline service station are
supposed to wash the windshield of each customer’s
car.
Janet, who services 20% of all cars, fails to wash the
windshield one time in 20
Tom, who services 60% of all cars, fails to wash the
windshield one time in 10
Georgia, who services 15% of all cars, fails to wash the
windshield one time in 10
Peter, who services 5% of all cars, fails to wash the
windshield one time in 20.
Conditional Probability
If a customer later complained that her
windshield was not washed, what is the
probability that her car was serviced by
Janet?
Solution
All these are mutually
exclusive events.
Pr[windshield was not
washed|Janet
serviced the car]
=
(.2)(.05)
(.2)(.05) (.6)(.1)  (.15)(.1) (.05)(.05)
 .114