Slide set 5
Stat 330 (Spring 2015)
Last update: January 13, 2015
Stat 330 (Spring 2015): slide set 5
Conditional Probability
Example 1: A box has 5 computer chips. Two are defective. Two chips are
selected from the box, one at a time.
1. Compute the probability that the second chip is defective.
Again common sense tells us that P ( a chip is defective) = 25 .
Using probability theory, we have
|Ω| = # of ways to draw 2 chips (ordered) = (5)(4)=20
Define event A = the second chip is defective
|A| = # of outcomes with second chip defective= (4)(2)=8
8
|A|
=
= .4.
P ( second chip is defective) =
|Ω| 20
2. If we know that the first chip drawn is good, what is the probability that
the second chip is defective?
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Stat 330 (Spring 2015): slide set 5
Conditional Probability (continued)
When we obtain additional information about a probability experiment, we
want to reassess the probabilities of events given the new information.
Definition: The conditional probability of an event A given an event B is
P (A|B) =
P (A ∩ B)
,
P (B)
provided P (B) 6= 0.
The above definition makes sense because the conditional probability of A
given B is the fraction of outcomes in B that are also in A.
An important implication of the definition is that
P (A ∩ B) = P (A|B)P (B) = P (B|A)P (A).
Knowing two of the above probabilities allow us to compute the third.
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Stat 330 (Spring 2015): slide set 5
Conditional Probability (continued)
Compute the probability that the second chip is defective given that the
first chip is good using the above definition.
Let A=the first chip is good; B = the second chip is defective.
Now we need to compute P (B|A)
First, we have that P (A) =
3
5
The event A ∩ B is first chip is good and the second chip is defective.
The # of outcomes with first chip is good and the second chip is defective
= (3)(2)
Thus we have that P (A ∩ B) =
Thus
P (B|A) =
6
20
P (A ∩ B) 6/20
=
= .5
P (A)
3/5
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Stat 330 (Spring 2015): slide set 5
More computer chips...
Example 2: A box has 500 computer chips with a speed of 400 Mhz and
500 with a speed of 500 Mhz. The numbers of good (G) and defective (D)
chips at the two different speeds are as shown below.
G
D
400 Mhz
480
20
500
500 Mhz
490
10
500
970
30
Total=1000
1. We select a chip at random and observe its speed. What is the probability
that the chip is defective given that its speed is 400 Mhz?
Drawing a chip at random has the following probabilities:
P (D) = 0.03
P (G) = 0.97
check: these two must sum to 1.
P (400mHz) = 0.5 P (500mHz) = 0.5 check: these two must sum to 1, too.
P (D and 400mHz) = 20/1000 = 0.02
P (D and 500mHz) = 10/1000 = 0.01
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Stat 330 (Spring 2015): slide set 5
Example 2 continued...
Suppose now, that I have the partial information that the chip selected is a
400 mHz chip.
2. What is now the probability that it is defective?
Using the definition of conditional probability, we get
P ( chip is D| chip is 400mHz) =
=
P ( chip is D and chip is 400mHz)
P ( chip is 400mHz)
0.02
= 0.04.
0.5
i.e. knowing the speed of the chip influences our probability assignment to
whether the chip is defective or not.
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Stat 330 (Spring 2015): slide set 5
Independence of Events
Sometimes, knowledge that B occurred does not change our assessment of
the P (A). Let’s say I toss a fair coin. I tell you that I got a tail. I then give
you the coin to toss. Does the knowledge that I got a tail affect what you
think the chance is that you will get a head?
Intuitively, two events A and B are independent if the event B does not
have any influence on the probability that A happens (and vice versa).
Mathematically, independence of two events is defined as follows:
Definition: Two events A and B are said to be independent, if
P (A ∩ B) = P (A) · P (B)
(Result: If P (B) 6= 0, then A and B are independent P (A|B) = P (A). )
Example: An alternative model for logging on to the AOL network using
dial-up.
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Stat 330 (Spring 2015): slide set 5
Independence Example
Example: Suppose I log on to AOL using dial-up. I connect successfully
if and only if the phone number works and the AOL network works. The
probability that the phone works is .9, and the probability that the network
works is .6. Suppose that the status of the phone line and the status of
the AOL network are independent. What is the probability that I connect
successfully?
Let A =“phone number works” and B = “AOL network works”; we need
P (A ∩ B).
Since we are told that A and B are independent, we know that
P (A ∩ B) = P (A) · P (B) = (.9)(.6) = .54
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Stat 330 (Spring 2015): slide set 5
Warning: independence and disjointedness are two very different concepts!
Disjointedness:
Independence:
If A and B are disjoint,
their intersection is empty,
has therefore probability 0:
If A and B are independent
events, the probability of
their intersection can be
computed as the product of
their individual probabilities:
P (A ∩ B) = P (A) · P (B)
The probability for the
intersection is zero only if A
or B is empty.
P (A ∩ B) = P (∅) = 0.
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Stat 330 (Spring 2015): slide set 5
On Systems in Series, Systems in Parallel, and Reliability
• A parallel system consists of k components c1, . . . , ck arranged in such
a way that the system works if and only if at least one of the k components
functions properly.
• A serial system consists of k components c1, . . . , ck arranged in such
a way that the system works if an only if all of the components function
properly.
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Stat 330 (Spring 2015): slide set 5
On Systems and Reliability (continued)
• The system consisting of the AOL network and the phone line is an
example of a serial system.
• The reliability of a system is the probability that the system works.
• For example, the reliability of the system consisting of the AOL network
and the phone line (if the two are independent) is .54.
• We can also construct larger systems with sub-systems that are connected
in series and/or in parallel.
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Stat 330 (Spring 2015): slide set 5
Example: Parallel system with k mutually independent
components
Let c1, . . . , ck denote the k components in a parallel system. Assume the
k components operate independently, and P (cj works ) = pj . What is the
reliability of the system?
P ( system works) = P ( at least one component works)
= 1 − P ( all components fail )
= 1 − P (c1 fails and c2 fails . . . and ck fails )
= 1−
k
Y
P (cj fails)
j=1
= 1−
k
Y
(1 − pj ).
j=1
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Stat 330 (Spring 2015): slide set 5
Example: System in series with k mutually independent
components
Let c1, . . . , ck denote the k components in a system. Assume the
k components are connected in series, operate independently, and
P (cj works ) = pj . What is the reliability of the system?
P ( system works) = P ( all components work)
=
k
Y
P (cj works)
j=1
=
k
Y
pj .
j=1
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Stat 330 (Spring 2015): slide set 5
Reliability of Systems (continued)
Example 2.17:Each component in the system shown below is opearable with
probability 0.92 independently of other components. Calculate the reliability
2.
The components D and
of the system:
E connected in parallel can be
replaced by component G that
operates with probability
P (D∪E) = 1−(1−0.92)2= 0.9936
1. The upper link A-B works if
both A and B work. Thus we can
replace this link with a component
F that operates with probability
P (A ∩ B) = (0.92)2 = 0.8464
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Stat 330 (Spring 2015): slide set 5
Example 2.17 (continued)
3. The components C and G connected in series so they can be replaced by
a component H that operates with probability
P (C ∩ G) = (0.92)(0.9936) = 0.9141
4. Lastly, the components F and H are in parallel so the reliability of the
system is
P (F ∪ H) = 1 − (1 − 0.8424)(1 − 0.9141) = 0.9868
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