Transcript Introduction to probability (3)

Introduction to probability (3)
• Definition: - The probability of an event
A is the sum of the weights of all
sample point in A therefore
0  P( A)  1
P ( )  0
P(S )  1
If A1,A2,…..,An is a sequence of mutually
exclusive events then
P( A1UA2 ,.........,UAn )  P( A1 )  P( A2 )  ........  P( An )
Introduction to probability (3)
• Rule: If an experiment can result in any
one of N different equally likely
outcomes and if exactly n of these
outcomes correspondent to event A,
then the probability of event A is
n
P ( A) 
N
Introduction to probability (3)
• Examples
1. A balanced coin tossed twice what is
the probability that at least 1 head
occurs.
• Solution: The number of ways for this
experiment is (2 )  4 ways. Then the
sample space is S  {HH , HT , TH , TT } , N =4
• A  {HH , HT , TH } The event of at least 1
head occurring. n=3.
2
n 3
P( A) 

N 4
Introduction to probability (3)
2. A die is loaded in such a way that an even
number is twice as likely to occur as an
odd number. If (E) is event that the number
less than 4 occurs in a single toss of the
die.
1. Find P(E)
• The sample space is S  {1,2,3,4,5,6}
•
Now we assign a probability of w to each
odd number and a probability of 2w to each
even number.
Then S  {w,2w, w,2w, w,2w}, N= 9w.
Introduction to probability (3)
•
•
E= {1, 2, 3} < 4
P (E) = {w, 2w, w}
P( E ) 
1 2 1 4
  
9 9 9 9
b. Let A be the event that an even number
turns up and let B be the event that a
number divisive by 3 occurs. Find
P ( A  B ) and
.
P( A  B)
Introduction to probability (3)
• Solution: Event A is {2, 4, 6}, event B is
{3, 6}.
P( A  B)  {2,3,4,6}, P(A  B)  {6}
2 1 2 2
7
P( A  B)     
9 9 9 9
9
2
P( A  B) 
9
Introduction to probability (3)
3. A statistics class for engineers consists of
25 industrial, 10 mechanics, 10 electrics an
8 civil engineering students. If a person is
randomly selected by instructor to answer
the question. Find the probability that the
student chosen is:
a) An industrial engineering major.
b) Civil engineering or an electrical engineering.
c) Mechanics and electrical engineering.
Introduction to probability (3)
• Solution:
•
Let I: denotes for industrial engineering.
•
M: denotes for mechanics engineering.
•
E: denotes for electrical engineering.
•
C: denotes for civil engineering.
N:= sum of student at class room.
N = 25 + 10 + 10 + 8 = 53 students.
a. Since 25 of 53 students are majoring industrial
engineering. Then the probability of event I selecting
an industrial engineering major at random is:
Introduction to probability (3)
1.
P( I ) 
2.
P(C ) 
25
53
8
10
, P(E) 
53
53
Then the probability a civil engineering or electrical
engineering is:
P(C  E )  P(C )  P( E ) 
3. ( M  E )  
P( M  E )  0
8 10 18


53 53 53
Introduction to probability (3)
4. In a poker hand consisting of 5 cards,
find the probability of holding 2 aces
and 3 Jacks.
• Solution: The total number of 5 cards
hands all of which are equal is
 52 
52!
52  51  50  49  48  47!
N    

 2598960
5
47
!

5
!
47
!

120
 
Introduction to probability (3)
• And
5
5! 5  4  3!
n1 for aces    
 10
 2  3!2! 3!2
 5
5! 5  4  3!
n2 for Jaks    
 10
 3  3!2! 3!2
Now there are
n  n1  n2  10 10  100
hands with 2 aces and 3 Jacks
Introduction to probability (3)
• Then the probability of getting 2 aces
and 3 Jacks in a 5 cards poker P(C) is
100
P(C ) 
 3.85  10 5
2598960