Download Introduction to probability (4)

Introduction to probability (4)
• Theorems:
P( A  B )  P( A)  P( B)
P( A )  1  P( A)
P( A)  P( A )  P(S )  1
P( A  B)  P( A)  P( B)  P( A  B)
Introduction to probability (4)
• Example: Two dice are tossed; find the
probability of getting an even number
on the first die or a total number of 8.
• Solution: S  (6 2 )  36
• A: Getting an even number on the first
die.
• B: The sum of the options obtained on
the two dice 8.
Introduction to probability (4)
A : {(2,4,6)  (1,2,3,4,5,6)}  n( A)  3  6  18
B : {( 2,6), (6,2), (3,5), (5,3), (4,4)}  n( B)  5
( A  B) : {( 2,6), (6,2), (4,4)}  n( A  B)  3
18 5
3 20
 P( A  B)  P( A)  P( B)  P( A  B) 



36 36 36 36
Introduction to probability (4)
• Example: If the probability that an
automobile machine will serve 3, 4, 5, 6,
7 or 8 or more cars on any given
workday are respectively: 0.12, 0.19,
0.28, 0.24, 0.10 and 0.07. What is the
probability that it will be serve at least 5
cars on next day at work.
• Solution: Let E be the event that at
least 5 cars are served
Introduction to probability (4)
 P( E )  1  P( E ) where E is the event that fewer
than 5 cars are served since :
P( E )  P(3)  P(4)  0.12  0.19  0.31
 P( E )  1  0.31  0.69
Conditional Probability
• The probability of an event B occurring
when it is known that some event A has
occurred is called a conditional
probability and is denoted by P ( B | A)
• and it can pronounced as “ The
probability of B given A”.
Conditional Probability
• Example: B is an event of getting a
perfect square when a die is tossed the
die is constructed so that the even
numbers are twice as likely to occur as
the odd numbers. Find the probability
that B occurs relative to the space A
and A is the number greater 3.
Conditional Probability
• Solution:
A : {4,5,6}  {2 w,1w,2 w}  5w
B : {4}  2 w
2
P( B | A) 
5
Conditional Probability
• Definition:
P( B | A) 
P( A  B)
, P( A)  0
P( A)
By the above definition we can solve the above
question by another way as:
S : {1,2,3,4,5,6}  9 w
A : {4,5,6}  {2 w,1w,2 w}  5w
B : {4}  2 w
P ( A) 
A  B  {4}  2w
P( A  B) 
5
9
2
9
2
P( A  B)
2 9
2
 P( B | A) 
 9   
5
P( A)
9 5
5
9
Conditional Probability
• Example:
Employed Unemployed
Total
Male
460
40
500
Female
140
260
400
Total
600
300
900
If we select one person from above table and the
event is:
M: a man is chosen.
E: the once chosen is employed
Conditional Probability
•
Solution: We can solve the above
problem by two methods as:
1. Directly from table as:
460 23
P( M | E ) 

600 30
Conditional Probability
2. Or
number( E  M )
number( E  M )
P( E  M )
nS
P( M | E ) 


n( E )
number ( E )
P( E )
n( S )
460 23
P( E  M ) 

900 45
600 2
P( E ) 

900 3
23
23 3
69
23
45
P( M | E ) 




2
45 2
90
30
3
Conditional Probability
• Example: The listed table is the number
of contaminated wafer:
No. Cont.
Center Edge Total
0
0.3
0.1
0.4
1
2
3
4
0.15
0.1
0.06
0.04
0.05
0.05
0.04
0.01
0.2
0.15
0.1
0.05
5 and more
Total
0.07
0.72
0.03
0.28
0.1
1
Conditional Probability
• Assume that one wafer is selected at
random from this set let A denotes the
event that a wafer contains four or
more particles and let B denotes the
event that a wafer is from a center.
• Find: P( A)
P( A)  (0.04  0.01  0.07  0.03)  0.15
Conditional Probability
P(B)
P( B) 
(03  0.15  0.1  0.06  0.04  0.07) 0.72

 0.72
1
1
P( A  B)
P( A  B)  0.04  0.07  0.11
P( A | B)
P( A | B) 
P( A  B) 0.11

P( B)
0.72
Conditional Probability
P ( B | A)
P( A  B) 0.11
P( B | A) 

P( A)
0.15
P( A  B)
P( A  B)  P( A)  P( B)  P( A  B)
 0.15  0.72  0.11  0.76