Download FM Lial 9th 7.4 Notes Sp10

7.4 Notes O’Brien Sp10
FM 9th ed Lial
7.4
Basic Concepts of Probability
I.
Important Rules and Properties of Probability
A.
The Union Rule for Probability
For any events E and F from a sample space S,
P(E U F) = P(E) + P(F) – P(E ∩ F)
We can think of the Union Rule as a formula with four variables. If we know the value of any three
of the variables, we can solve for the fourth. For example, if we know P(E), P(F), and P(E U F), we
can find the probability of E ∩ F as follows: P(E ∩ F) = P(E) + P(F) – P(E U F).
Note: For mutually exclusive events, E ∩ F = Ø and P(E ∩ F) = 0, therefore P(E U F) = P(E) + P(F).
B.
Complement Rule
P(E) = 1 – PE
C.
and
PE = 1 – P(E)
P(E) + PE = 1
and
Properties of Probability
Let S be a sample space consisting of n distinct outcomes, s1, s2, s3, , sn.
1.
The probability of each outcome is a number between 0 and 1.
0 ≤ p1 ≤ 1, 0 ≤ p2 ≤ 1, 0 ≤ p3 ≤ 1, …, 0 ≤ pn ≤ 1
2.
The sum of the probabilities of all possible outcomes is 1.
p1 + p2 + p3 + … + pn = 1
D.
Theoretical versus Empirical Probabilities
Theoretical probabilities are based on deductive reasoning alone.
Empirical probabilities are based on observation or experimentation.
Example 1
The numbers 1, 2, 3, 4, and 5 are written on slips of paper. Two slips are drawn at random, one at a time,
without replacement. Find the probability of the following events.
a.
[20]
Both numbers are even
Assuming the order in which the numbers are drawn matters, the sample space for the experiment is:
S = {(1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3),
(4, 5), (5, 1), (5, 2), (5, 3), (5, 4)}; n(S) = 20
E: Both numbers are even;
b.
E = {(2, 4), (4, 2)};
nE
PE  nS =
n(E) = 2;
2
20
=
1
10
One of the numbers is even OR one of the numbers is greater than 3
Let’s use the Union Rule to find this probability.
E: One of the numbers is even;
E = {(1, 2), (1, 4), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4)};
P(E) = 12
= 35
20
n(E) = 12;
F: One of the numbers is greater than 3;
F = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3)};
P(F) = 12
= 35
20
n(F) = 12;
E ∩ F: One of the numbers is even AND one of the numbers is greater than 3;
E ∩ F = {(2, 4), (2, 5), (4, 2), (4, 5), (5, 2), (5, 4)}; n(E ∩ F) = 6; P(E ∩ F) =
Therefore, P(E U F) = P(E) + P(F) – P(E ∩ F) =
3
5
+
3
5
–
3
10
=
6
10
+
6
10
–
6 = 3
20
10
3 = 9
10
10
1
7.4 Notes O’Brien Sp10
FM 9th ed Lial
c.
The sum is 5 OR the second number is 2
This time let’s define E as our final desired outcome, “the sum is 5 OR the second number is 2”.
E = {(1, 2), (1, 4), (2, 3), (3, 2), (4, 1), (4, 2), (5, 2)};
nE
PE  nS =
n(E) = 7;
7
20
By including the outcomes where “the sum is 5 AND the second number is 2” only once in E, we did not
have to use the Union Rule. This is possible when you can list the elements in the event space or when
you can determine n(E ∩ F) or P(E ∩ F).
II.
Finding Probabilities Using a Venn Diagram
One way to find specified probabilities is to use a Venn Diagram.
Example 2
Suppose P(E) = .26, P(F) = .41, and P(E ∩ F) = .16. Use a Venn Diagram to find the following.
a.
P(E U F)
P(E' ∩ F)
b.
P(E) = a + b = .26
c.
P(E ∩ F')
d.
[21]
P(E' U F')
P(E ∩ F) = b = .16
P(F) = b + c = .41
Therefore, a = .26 – .16 = .10 and c = .41 – .16 = .25 and d = 1 – .10 – .16 – .25 = .49
a.
P(E U F) = a + b + c = .10 + .16 + .25 = .51
b.
P(E' ∩ F) = c = .25
c.
P(E ∩ F') = a = .10
d.
P(E') = 1 – .26 = .74; P(F') = 1 – .41 = .59; P(E' ∩ F') = d = .49
P(E' U F') = P(E') + P(F') – P(E' ∩ F') = .74 + .59 – .49 = .84
III.
Probabilities with Playing Cards
A standard deck of cards has a total of 52 cards.
The cards are divided into four suits: clubs, diamonds, hearts, and spades.
Clubs and spades are black; diamonds and hearts are red.
Each suit has 13 cards: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king.
Jacks, queens, and kings are called face cards.
Hearts
Clubs
Diamonds
Spades
A♥
A♣
A♦
A♠
2♥
2♣
2♦
2♠
3♥
3♣
3♦
3♠
4♥
4♣
4♦
4♠
5♥
5♣
5♦
5♠
6♥
6♣
6♦
6♠
7♥
7♣
7♦
7♠
8♥
8♣
8♦
8♠
9♥
9♣
9♦
9♠
10♥
10♣
10♦
10♠
J♥
J♣
J♦
J♠
Q♥
Q♣
Q♦
Q♠
K♥
K♣
K♦
K♠
Example 3
One card is drawn from an ordinary deck of 52 cards. Find the probabilities of drawing the following cards.
a.
Less than a 4 (count aces as one)
E = { A♥, 2♥, 3♥, A♣, 2♣, 3♣, A♦, 2♦, 3♦, A♠, 2♠, 3♠};
b.
[16]
n(E) = 12;
P(E) =
12
52
3
 13
A diamond or a 7
E = { A♦, 2♦, 3♦, 4♦, 5♦, 6♦, 7♦, 8♦, 9♦, 10♦, J♦, Q♦, K♦, 7♥, 7♣, 7♠};
n(E) = 16;
P(E) =
16
52
4
 13
2
7.4 Notes O’Brien Sp10
FM 9th ed Lial
c.
A black card or an ace
E = {A♣, 2♣, 3♣, 4♣, 5♣, 6♣, 7♣, 8♣, 9♣, 10♣, J♣, Q♣, K♣, A♠, 2♠, 3♠, 4♠, 5♠, 6♠, 7♠,
 7
8♠, 9♠, 10♠, J♠, Q♠, K♠, A♥, A♦}; n(E) = 28; P(E) = 28
52 13
d.
A heart or a jack
E = {A♥, 2♥, 3♥, 4♥, 5♥, 6♥, 7♥, 8♥, 9♥, 10♥, J♥, Q♥, K♥, J♣, J♦, J♠};
e.
n(E) = 16;
P(E) =
16
52
4
 13
A red card or a face card
E = {A♥, 2♥, 3♥, 4♥, 5♥, 6♥, 7♥, 8♥, 9♥, 10♥, J♥, Q♥, K♥, A♦, 2♦, 3♦, 4♦, 5♦, 6♦, 7♦,
8
 13
8♦, 9♦, 10♦, J♦, Q♦, K♦, J♣, Q♣, K♣, J♠, Q♠, K♠}; n(E) = 32; P(E) = 32
52
IV.
Probabilities with Dice
A.
Rolling a Single Die
An ordinary die is a cube whose six different faces show the following number of dots: 1, 2, 3, 4, 5, 6.
The sample space for rolling a single die is S = {1, 2, 3, 4, 5, 6}.
If the die is fair, then any one of the faces is equally likely to come up when the die is rolled; i.e.,
P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 61 .
B.
Rolling Two Dice
1.
If two dice are rolled, one black and one white, there are thirty-six possible outcomes.
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
If the two dice are fair, then the probability of each possible out come is the same,
2.
1
36
.
If we consider the outcome of each trial to be the SUM of the dots on the tops of the two dice,
there are 11 possible outcomes, S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. These outcomes are
not equally likely since there is one way to roll a 2, two ways to roll a 3, three ways to roll a 4,
and so on, as shown in the table below.
sum
2
3
4
5
6
7
8
9
10
11
12
# ways to roll
1
2
3
4
5
6
5
4
3
2
1
probability
1
36
2
36
3
36
4
36
5
36
6
36
5
36
4
36
3
36
2
36
1
36
Note: A table containing a row or column listing the possible outcomes of an experiment and a
second row or column listing the number of ways of obtaining each outcome (the frequency
of the outcome) is called a frequency distribution.
A table containing a row or column listing the possible outcomes of an experiment and a
second row or column listing the probability of each outcome is called a relative frequency
or probability distribution. The relative frequency of an outcome is the same as its
probability.
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7.4 Notes O’Brien Sp10
FM 9th ed Lial
Example 4
Two dice are rolled. Find the probabilities of rolling the given sums.

a.
9 or more
b.
Less than 7
c.
Between 5 and 8 (exclusive)
V.
Odds
A.

E = {9, 10, 11, 12}


E = {2, 3, 4, 5, 6}

4
36
1
= 36
P(E) =
P(E)

E = {6, 7}
[10]
3
36
2
+ 36
+
2 + 1 = 10
36
36
36
3
5
4
+ 36 + 36 + 36
5 + 6 = 11
36
36
36
+
P(E) =
5
18
= 15
36
=
=
5
12
Finding Odds from Probabilities
The Odds in Favor of an event E are defined as the ratio of P(E) to PE .
O(E) =
P(E)
PE 
, where PE ≠ 0.
If P(E) = A and PE = B, the odds in favor of E are written as A to B
or
A : B.
The Odds Against an event E are defined as the ratio of PE to P(E).
PE
OE  PE , where P(E) ≠ 0.
If P(E) = A and PE = B, the odds against E are written as B to A
or
B : A.
Example 5
Two dice are rolled. Find the following odds.
a.
odds for rolling a sum of 8
P(E) =
b.
5
36
P(E') = 1 –
5
36

=
E: sum of 8
31
36
PE 
=
PE 
O(E) =

odds for rolling a sum of 3 or a sum of 7
5
36
31
36
=
5  36
36 31
c.
2
9
=
7
9
;
PE 
=
PE 
O(E) =

odd against rolling a sum less than 6
P(E) =
1
36
2
36
+
3
36
+
OE  
PE 
=
PE
13
18
5
18
=
13  18
18
5
B.
+
4
36
=
10
36
=
13
5
=
2
9
7
9
5
18
;

=
O(E) = 5 to 31
E: sum of 3 or sum of 7
Since these are mutually exclusive events, P(3 or 7) = P(3) + P(7) =
PE = 1 –

5
= 31
29
9 7

= 72
5
18
6
36
+
=
8
36
=
2
9
;
O(E) = 2 to 7
E: sum less than 6
PE = 1 –
2
36
=

E: {2, 3, 4, 5}
13
18
OE = 13 to 5
Finding Probabilities from Odds
If the odds in favor of an event are m to n, then
m
n
PE  
PE  
and
mn
mn
Example 6
If the odds that a given candidate will win an election are 3 to 2, what is the probability that the candidate
will lose?
E: candidate wins;
O(E) = 3 to 2

m = 3;
n = 2;
E : candidate loses;
PE  
[73]
2
2
n
=
=
32
5
mn
4