#### Transcript exactly one defective

```Chapter 4: Probability
(Cont.)
In this handout:
• Total probability rule
• Bayes’ rule
• Random sampling from finite population
• Rule of combinations
Partitions
• Definition: A collection of events {S1, S2, …, Sn} is
a partition of a sample space S if
1. S = S1  S2  …  Sn
2. S1, S2, …, Sn are mutually exclusive events.
• Example:
Recall the example of rolling two dice.
Define S2={sum = 2}, S3={sum = 3}, …, S12={sum = 12}.
Then {S2, S3, …, S12} is a partition of the sample space.
Total probability rule
Assume the set of events {S1, S2, …, Sn} is a partition of a
sample space S. Assume P(Si) > 0 for every i, 1 ≤ i ≤ n.
Then for any event A,
n
P ( A)   P ( S i )  P ( A | S i )
i 1
S1
AS1
S2
S3
AS2
AS3
S4
AS4
Total probability rule
Example: A diagnostic test for a certain disease is known to be 95%
accurate. It is also known from previous data that only 1% of the
population has the disease. What is the probability that a person
chosen at random will be tested positive?
Solution: Let
T+ denote the event that a person is tested positive;
T- denote the event that a person is tested negative;
D denote the event that a person has the disease.
Then
P( D)  0.01, P( D )  0.99, P(T  | D)  0.95, P(T  | D )  0.95
Applying the formula of total probability:
P(T  )  P(T  | D)  P( D)  P(T  | D )  P( D )
 0.95  0.01  0.05  0.99  0.059
Bayes’ rule
Assume the set of events {S1, S2, …, Sn} is a partition of a
sample space S. Assume P(Si) > 0 for every i, 1 ≤ i ≤ n.
Fix any event A. Then for any given j, 1 ≤ j ≤ n,
P( S j | A) 
P( S j )  P( A | S j )
n
 P( S )  P( A | S )
i 1
i
i
Example (cont.): If a person tested positive then what is the
probability he has the disease?

P
(
T
| D)  P( D)

P( D | T ) 
P(T  | D)  P ( D)  P (T  | D )  P( D )
0.95  0.01

 0.16
0.95  0.01  0.05  0.99
Random sampling from finite population
When sampling from a large population, listing all the possible choices
becomes a tedious job. Then a counting rule is very useful:
Note: C(N,r) is another way to denote “N choose r”.
Example: Randomly select 4 members from a group of 11 students to
work on a project. How many distinct 4-person teams can be
chosen?
11 1110  9  8
  
 330
 4  1 2  3  4
Random sampling from finite population
Example(cont.):
 The probability that students A,B,C,D are chosen to work on the
project is 1/330.
 Suppose the group consists of 5 juniors and 6 seniors. How many
samples of 4 have exactly 3 juniors?
Think of selecting a sample as a 2-step process:
• 1 senior can be chosen 6 different ways;
• 3 juniors can be chosen C(5,3) different ways.
Thus, the whole sample can be chosen 6∙C(5,3) different ways.
Examples on Combinations
 Suppose that 3 cars in a production run of 40 are defective.
A sample of 4 is to be selected to be checked for defects.
Questions:
1) How many different samples can be chosen?
2) How many samples will contain
exactly one defective car?
3) What is the probability that a randomly chosen sample
will contain exactly one defective car?
4) How many samples will contain
at least one defective car?
Solution:
1) C(40, 4) = (40∙39∙38∙37) / (1∙2∙3∙4) = 91,390
Examples on Combinations
2) How many samples will contain
exactly one defective car?
Think of selecting a sample as a 2-step process:
Step 1: Choose the defective cars;
Step 2: Choose the good cars.
There are C(3,1) ways to choose 1 defective car.
There are C(37,3) ways to choose 3 good cars.
Then the number of samples containing exactly 1 defective car is
C(3,1) ∙ C(37,3) = 3∙(37∙36∙35) / (1∙2∙3) = 23,310
3) What is the probability that a randomly chosen sample
will contain exactly one defective car?
The probability = 23,310 / 91,390 = .255
Examples on Combinations
4) How many samples will contain at least one defective car?
There are 2 ways to answer this question,
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