Stat 330 (Spring 2015): slide set 5
Knowing two of the above probabilities allow us to compute the third.
P (A ∩ B) = P (A|B)P (B) = P (B|A)P (A).
An important implication of the definition is that
2
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.
provided P (B) = 0.
P (A ∩ B)
P (A|B) =
,
P (B)
Definition: The conditional probability of an event A given an event B is
When we obtain additional information about a probability experiment, we
want to reassess the probabilities of events given the new information.
Conditional Probability (continued)
Last update: January 13, 2015
Stat 330 (Spring 2015)
Slide set 5
Stat 330 (Spring 2015): slide set 5
1
Stat 330 (Spring 2015): slide set 5
3
5
Thus
P (B|A) =
6
20
P (A ∩ B) 6/20
=
= .5
P (A)
3/5
Thus we have that P (A ∩ B) =
3
The # of outcomes with first chip is good and the second chip is defective
= (3)(2)
The event A ∩ B is first chip is good and the second chip is defective.
First, we have that P (A) =
Now we need to compute P (B|A)
Let A=the first chip is good; B = the second chip is defective.
Compute the probability that the second chip is defective given that the
first chip is good using the above definition.
Conditional Probability (continued)
P ( second chip is defective) =
|A|
8
=
= .4.
|Ω| 20
2. If we know that the first chip drawn is good, what is the probability that
the second chip is defective?
Define event A = the second chip is defective
|A| = # of outcomes with second chip defective= (4)(2)=8
|Ω| = # of ways to draw 2 chips (ordered) = (5)(4)=20
Using probability theory, we have
Again common sense tells us that P ( a chip is defective) = 25 .
1. Compute the probability that the second chip is defective.
Example 1: A box has 5 computer chips. Two are defective. Two chips are
selected from the box, one at a time.
Conditional Probability
Stat 330 (Spring 2015): slide set 5
500 Mhz
490
10
500
970
30
Total=1000
P ( chip is D and chip is 400mHz)
P ( chip is 400mHz)
0.02
= 0.04.
0.5
Definition: Two events A and B are said to be independent, if
Mathematically, independence of two events is defined as follows:
6
(Result: If P (B) = 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.
7
Since we are told that A and B are independent, we know that
P (A ∩ B) = P (A) · P (B) = (.9)(.6) = .54
Let A =“phone number works” and B = “AOL network works”; we need
P (A ∩ B).
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?
Independence Example
Stat 330 (Spring 2015): slide set 5
Stat 330 (Spring 2015): slide set 5
i.e. knowing the speed of the chip influences our probability assignment to
whether the chip is defective or not.
=
P ( chip is D| chip is 400mHz) =
Using the definition of conditional probability, we get
2. What is now the probability that it is defective?
5
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).
P (A ∩ B) = P (A) · P (B)
Stat 330 (Spring 2015): slide set 5
Suppose now, that I have the partial information that the chip selected is a
400 mHz chip.
Example 2 continued...
4
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?
Independence of Events
P (D and 400mHz) = 20/1000 = 0.02
P (D and 500mHz) = 10/1000 = 0.01
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.
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:
G
D
400 Mhz
480
20
500
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.
More computer chips...
10
• We can also construct larger systems with sub-systems that are connected
in series and/or in parallel.
• For example, the reliability of the system consisting of the AOL network
and the phone line (if the two are independent) is .54.
• The reliability of a system is the probability that the system works.
• The system consisting of the AOL network and the phone line is an
example of a serial system.
= 1−
= 1−
j=1
k
j=1
k
(1 − pj ).
P (cj fails)
= 1 − P (c1 fails and c2 fails . . . and ck fails )
= 1 − P ( all components fail )
P ( system works) = P ( at least one component works)
11
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?
Example: Parallel system with k mutually independent
components
Stat 330 (Spring 2015): slide set 5
Stat 330 (Spring 2015): slide set 5
On Systems and Reliability (continued)
9
• 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.
• 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.
8
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.
If A and B are disjoint,
their intersection is empty,
has therefore probability 0:
P (A ∩ B) = P (∅) = 0.
Independence:
Disjointedness:
Stat 330 (Spring 2015): slide set 5
On Systems in Series, Systems in Parallel, and Reliability
Stat 330 (Spring 2015): slide set 5
Warning: independence and disjointedness are two very different concepts!
Stat 330 (Spring 2015): slide set 5
j=1
k
j=1
pj .
P (cj works)
Stat 330 (Spring 2015): slide set 5
12
P (F ∪ H) = 1 − (1 − 0.8424)(1 − 0.9141) = 0.9868
14
4. Lastly, the components F and H are in parallel so the reliability of the
system is
P (C ∩ G) = (0.92)(0.9936) = 0.9141
3. The components C and G connected in series so they can be replaced by
a component H that operates with probability
Example 2.17 (continued)
=
=
k
P ( system works) = P ( all components work)
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?
Example: System in series with k mutually independent
components
Stat 330 (Spring 2015): slide set 5
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
13
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
Reliability of Systems (continued)