Probability, Lec 2

Lecture 2
Example (8): A coin is tossed twice. What is the

probability that at least one head occurs?
Solution: The sample space for this experiment is

S = {(H, H), (H, T), (T, H), (T, T)} , n(S)=4
If the coin is balanced, each of these outcomes would
be equally to occur.
If A represents the event of at least one head
occurring, then
A = {(H, H), (H, T), (T, H)}, n(A)=3 , and
n (A) 3
P(A)  
n (S) 4
Definition of bridge and poker
A deck of bridge cards consists of 52 cards arranged in four
suits of thirteen each . There are thirteen face values
(2, 3, 4, …, 10, jack, queen, king, ace ) in each suit .
The four suit are called
“ spades’’ , “clubs “ , “hearts” , “diamonds” .
The last two are red , the first two black . Cards of the same
face value are called of the same kind .For our purposes ,
playing bridge means distributing the cards to four players ,
to be called North , South , East , and West ( or N, S, E, W ,
for short ) so that each receives thirteen cards . Playing
poker , by definition , means selecting five cards out of the
pack .
Example (9):
A card is drawn from a standard deck. Find the probabilities of the
following events.
1. An ace (Book)
2. Getting a queen.
3. Getting a club.
4. Getting a number.
Solution: Note that n(S) = 52
1. A is the event of getting an ace . P(A) = n(A) / n(S) = 4 / 52 = 1 / 13
2. B is the event of getting a queen . P(B) = n(B) / n(S) = 4 / 52 = 1 / 13
3. C is the event of getting a club. P(C) = n(C) / n(S) = 13 / 52 = 1 / 4
4. D is the event of getting a number
P(D) = n(D) / n(S) = 40 / 52 = 10 / 13
Example (10) : Page 59
If an experiment has three possible and mutually exclusive outcomes A,
B, and C , check in each case whether the assignment of probabilities is
permissible :
1 1 1
(a) P(A)  , P(B)  and P(C)  3
3 3
(b) P(A) = 0.64 , P(B) = 0.38 , and P(C) = -0.02
(c) P(A) = 0.35 , P(B) = 0.52 , and P(C) = 0.26
(d) P(A) = 0.57 , P(B) = 0.24 , and P(C) = 0.19

Solution:
(a) The assignment of probabilities is permissible because
(i) the values are all on the interval from 0 to 1, and
1 1 1
(ii) their sum is   1
3 3 3
(b) The assignment is not permissible because P(C) is negative (< 0 )
(c) The assignment is not permissible because 0.35 + 0.52 + 0.26 =

1.13 > 1
(d) The assignment is permissible because

(i) the values are all on the interval from 0 to 1, and
(ii) their sum is 0.57 + 0.24 + 0.19 = 1
Some Simple Properties
From general properties of sets and the properties of definition
of probability we can derive some other useful properties of
probability
1- If A is an event and A is its complement , then

P(A) = 1- P( A)
2- If A and B are subsets of S then

P(A ∩ B ) = P(A) –P(A∩B)
3- If A and B are subsets of S then
P(A  B)  P(A)  P(B)  P(A  B)

4- If A and B are two events of S such that A  B, then
P (A) ≤ P (B)
5- De Morgan's laws:
(a) P(A  B)  P(A  B)  1  P(A  B)
(b)
P(A  B)  P(A  B)  1  P(A  B)
Example (11): The probability that a student passes
mathematics is 2/3, and the probability that he passes English
is 4/9. If the probability of passing both courses is 1/4, what is
the probability that the student will pass at least one of these
courses?
Solution: Let, M is the event "passing mathematics," and
E the event that "passing English", then
P(M  E)  P(M )  P(E)  P(M  E)  2  4  1  31

3 9 4 36
Example (12): What is the probability of getting a total of
7 or 11 when a pair of dice is tossed?
Solution: S = { (i , j ) : i , j = 1, 2, 3, 4, 5, 6 } , and
n(S) = 36 .
Let A be the event that 7 occurs and
B the event that 11 comes up.
Now
A = { (1,6),(2,5),(3,4) , (4,3) , (5,2) , (6,1) } ,
B = { (5,6) , (6,5) } ,
n(A) = 6 ,
n (A) 6 1
P( A )   
n (S) 36 6
so n(B) = 2 , so
n (B) 2 1
P(B)   
n (S) 36 18
The events A and B are mutually exclusive, since a total of 7

and 11 cannot both occur on the same toss.
Therefore,
P(A  B)  P(A)  P(B)  1  1  2

6 18 9
Example (13): Let A and B be events with
P(A)  1 , P(B)  83 , and P(A  B)  1

Find 2 4
P(A  B) , P(B) , P(A  B) , P(A  B) , P(A B)

Solution:
(i)
P(A  B)  P(A)  P(B)  P(A  B)  1  3  1  5
2 8 4 8
(ii)
P(B)  1  P(B)  1  83  5
8
(iii)
P(A  B)  P(A  B)  1  P(A  B)  1  85  83
(iv)
P(A  B)  P(A  B)  1  P(A  B)  1  14  43
(v)
P(A  B)  P(A)  P(A  B)  1
2  1
4  1
4
Example (14): A class consists of 30 students, 18 of them
study statistics, 10 study computer and 6 study both statistic
and computer. If a student is selected at random.
Calculate the probability that this student:
(a) is one of those studying at least one of the subjects.
(b) is one of those studying at most one of the subjects?
(c) does not study any of the subjects.
(d) is one of those studying only one subject.

Let S = the event that , the student study Statistics , and
C= the event that , the student study Computer
P(S) = 18/30 = 3/5 , P(C ) = 10/30 = 1/3 and P(S∩C )= 6/30 = 1/5
(a) P(S  C)  P(S)  P(C)  P(S  C)  53  13  15  15

11
(b) P(S  C)  P(S  C)  1  P(S  C)  1  15  4

5
(C) P(S  C)  P(S C)  1  P(A  B)  1  1511  154

(d ) P(S only) + P(C only) = P( S ∩ C ) + P( S ∩ C)
= [P(S)-P(S∩C) ] + [P(C)-P(S∩C) ]
= P(S) + P(C)-2P(S∩C)
312 8
5 3 5 15
Basic Theorems:
If A & B are two events from S for a random experiment then:
• The probability of at least one event occurs is
If A&B are disjoint then

.
• The probability of non of them to occur is
.
• The probability of non of them to happen together is
.
• The probability of A not occurring is
.
• The probability of only A to happen is

Probability, Lec 2

Uploaded by

Copyright:

Available Formats

Probability, Lec 2

Uploaded by

Document Information

Original Description:

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Probability, Lec 2

Uploaded by

Copyright:

Available Formats

Lecture 2

Example (8): A coin is tossed twice. What is the

Solution: The sample space for this experiment is

(c) P(A) = 0.35 , P(B) = 0.52 , and P(C) = 0.26

(d) P(A) = 0.57 , P(B) = 0.24 , and P(C) = 0.19

(b) The assignment is not permissible because P(C) is negative (< 0 )

(c) The assignment is not permissible because 0.35 + 0.52 + 0.26 =

(d) The assignment is permissible because

1- If A is an event and A is its complement , then

2- If A and B are subsets of S then

P(A  B)  P(A)  P(B)  P(A  B)

(a) P(A  B)  P(A  B)  1  P(A  B)

P(M  E)  P(M )  P(E)  P(M  E)  2  4  1  31

The events A and B are mutually exclusive, since a total of 7

P(A  B)  P(A)  P(B)  1  1  2

P(A)  1 , P(B)  83 , and P(A  B)  1

P(A  B) , P(B) , P(A  B) , P(A  B) , P(A B)

(b) is one of those studying at most one of the subjects?

(c) does not study any of the subjects.

(d) is one of those studying only one subject.

(a) P(S  C)  P(S)  P(C)  P(S  C)  53  13  15  15

(b) P(S  C)  P(S  C)  1  P(S  C)  1  15  4

(C) P(S  C)  P(S C)  1  P(A  B)  1  1511  154

• The probability of at least one event occurs is

If A&B are disjoint then

• The probability of non of them to occur is

You might also like