Download Unit VI (4.1 – 4.4): Probability

Probability
Essential Ideas for
The Nature of Probability
An event is an outcome from an experiment. An
experiment is any process that can be repeated in
which the results are uncertain.
The probability of an event is a measure of the
likelihood of its occurrence. The long-term proportion
with which a certain outcome is observed is the
probability of that outcome.
A probability model lists the different outcomes from
an experiment and their corresponding probabilities
To construct probability models, we must know the
sample space of the experiment. The sample space S
lists all the possible outcomes of an experiment.
For example, the sample space resulting from the
experiment of rolling a die is:
S = {1, 2, 3, 4, 5, 6}
Properties of Probabilities
For a sample space S:
The probability of each event in S occurring is between 0
and 1.
The sum of the probabilities of all events in S occurring is
equal to 1.
If an event is impossible, the probability of the event is 0.
If an event is a certainty, the probability of the event is 1.
There are three methods for determining
the probability of an event:
1.The classical (theoretical) method
2.The empirical (experimental) method
3.The subjective method
Classical Method
The classical method of computing probabilities
requires equally likely outcomes, meaning
that each event has the same probability of
occurring. The corresponding theoretical
probabilities are obtained by logical
reasoning according to stated definitions.
Computing Probability Using the Classical
Method
If an experiment has n equally likely simple
events and if the number of ways that an event
E can occur is s, then the probability of E,
denoted P(E), is:
s
P (E ) 
n
Where s = number of ways that E can occur and
n = Total number of possible outcomes
Examples of theoretical probabilities
Suppose that a single card is drawn from an
ordinary deck of 52 cards.
Then,
P(ace) = 4/52 = 1/13 ≈ 0.08
P(heart) = 13/52 = 0.25
P(face card) = 12/52 = 3/13 ≈ 0.23
Suppose you are beginning a game of Monopoly. What is
the probability that you land on a railroad on your first roll
of the dice? The sample space for rolling a pair of dice is
given below.
Since you roll a pair of dice in the game of Monopoly, you
must determine how many spaces there are to the first
railroad. (It is impossible to get to the other three railroads
on the first roll, because they are more than 12 spaces
away.)
You must move five spaces on your first roll of the dice to
land on the first railroad.
Looking back at the sample space, there are four ways to
obtain a combined roll of “5”, so the probability of landing
on a railroad on your first roll of the dice is:
P(five) = 4/36 = 1/9 = 0.11
Empirical Method
 Empirical probabilities are obtained from
experimental data.
 As an experiment is repeated more and more
times, the empirical probability will tend to
come closer and closer to the theoretical
probability of that event.
Computing Probability Using the Empirical Method
The probability of an event E is approximately
the number of times E is observed divided by
the number of repetitions of the experiment.
For example, suppose that in a certain study, 46
out of 155 people showed a certain kind of
behavior.
P(behavior shown) = 46/155 ≈0.30
This is an empirical probability because it was
arrived at through experimentation.
Law of Large Numbers
The theoretical probability of rolling a die and
obtaining a “3” is 1/6 because there are six
possible outcomes, each with an equal
chance of winning. If we roll a die 100 times
and obtain a “3” on twenty rolls, the empirical
( or experimental) probability is 20/100 =
1/5. As we continue to roll the die, the
empirical probability will tend to come closer
and closer to the theoretical probability of 1/6.
This is an example of the Law of Large
Numbers.
Subjective Method
Subjective probabilities are probabilities based
upon an educated guess.
For example, there is a 30% chance of rain
tomorrow.
Probabilities of Unions and Intersections
Let E and F be two events.
P(E U F) is the probability that either E or F
occurs
P(E ∩ F) is the probability that both E and F
occur.
Suppose that a single card is drawn from an
ordinary deck of card.
What is the probability that it is a two or a king?
P(two U king) = (4 + 4)/52 = 8/52 = 2/13 ≈ 0.15
What is the probability that it is a two or a heart?
P(two U heart) = (4 + 12)/52 = 16/52 = 4/13 ≈ 0.31
What is the probability that it is a two and a heart?
P(two ∩ heart) = 1/52 ≈ 0.02
What is the probability that it is a two and a king?
P(two ∩ king) = 0/52 = 0
Independent Events
Two events E and F are independent if the
occurrence of event E in a probability
experiment does not affect the probability of
event F.
Two events E and F are dependent if the
occurrence of event E in a probability
experiment does affects the probability of
event F.
Independent Events and Intersection
If events E and F are independent events,
then we can find the probability of an
intersection using the multiplication
property of a probability:
P(E and F) = P(E ∩ F) = P(E)·P(F)
Example: Suppose a coin is tossed and a
die is simultaneously rolled.
What is the probability of tossing a tail and
rolling a “4”?
P(tail ∩ 4) = (1/2)·(1/6) = 1/12
Independent Events and Union
If events E and F are independent events,
then we can find the probability of a union
using the addition property of
probability:
P(E or F) = P(E U F) = P(E) + P(F) – P(E ∩ F)
= P(E) + P(F) – P(E)·P(F)
Example: Suppose a coin is tossed and a
die is simultaneously rolled.
What is the probability of tossing a tail or rolling
a “4”?
P(tail U 4) = P(tail) + P(4) – P(tail ∩ 4)
= 1/2 + 1/6 – (1/2)(1/6) = 7/12 ≈ .58
Mutually Exclusive Events
If events E and F have no simple events in common or
cannot occur simultaneously, they are said to be
mutually exclusive or disjoint. The addition rule for
mutually exclusive events simplifies to:
P(E U F) = P(E) + P(F)
because P(E ∩ F) = 0 if the events are mutually
exclusive.
Note: Mutually exclusive events and independent
events are not synonymous!
Complementary Probabilities
Let
s = number of ways an event can occur (successes)
f = number of ways the event cannot occur (failures)
n = total number of possible outcomes
so that (s + f) = n
Then, the probability that event E occurs is P(E) = s/n
The probability that event E does not occur is P (E ) = f/n
so that P(E) + P (E ) = 1
Two important properties of complements are:
P (E )  1  P (E )
P (E )  1  P (E )
What is the probability of obtaining at least one
head in three flips of a coin?
We could answer this with a tree diagram, but it is
easier to use the complement along with the
multiplication rule.
The complement of this problem is to determine the
probability of obtaining no heads in three flips of a
coin. This is the same thing as determining the
probability of obtaining all tails in three flips of a coin.
We can use the multiplication property of a probability
to answer this question:
P(tail ∩ tail ∩ tail) = (1/2)·(1/2)·(1/2) = 1/8
So, the probability of obtaining at least one head is:
1 – 1/8 = 7/8
Conditional Probabilities
A conditional probability is a probability of an
event given that another event has occurred.
We denote this by:
P (E F )
Read this as: “probability of E given F”
A conditional probability will alter the sample
space of E.
A single card is drawn from a standard deck of cards.
Find the following conditional probabilities.
P (facecard jack )
The probability of drawing a face card given that
the card drawn is a jack is 1 (since all jacks are
face cards).
P ( jack facecard )
The probability of drawing a jack given that the
card drawn is a face card is 4/12 = 1/3 because
there are 12 face cards in a standard deck and 4 of
them are jacks.
Drawing with Replacement
Drawing with replacement means choosing the first
item, noting the result, and then replacing the item
back into the sample space before selecting the
second item. With replacement, the events are
independent.
Find the probability of drawing a spade on the first draw
and a heart on the second draw with replacement.
13 13
P (S1  H2 )  P (S1 )  P (H2 ) 

 1/16  .0625
52 52
Drawing without Replacement
Drawing without replacement means choosing the first
item, noting the result, and then selecting a second
item without replacing the first item.
Find the probability of drawing a spade on the first draw
and a heart on the second draw without replacement.
13 13
P (S1  H2 )  P (S1 )  P (H2 S1 ) 
  13 / 204  .064
52 51
Expected Value / Expectation
A game is said to be fair if the expected
value is 0. If the expected value is positive,
then the game is in your favor. If the expected
value is negative, then the game is not in
your favor. The expected value (or your
expectation) of winning is computed using the
following formulas:
Expected Value Formulas
If there is no cost to play:
Expectation = (Amt. to win) x (Prob. of winning)
If there is an “up front” cost for playing the game, then:
Expectation = (Amt. to win) x (Prob. of win) – Cost of playing
If there is a “leave your money on the table” cost, then:
Expectation = (Amt. to win) x (Prob. of win) – (Cost to
play)(Prob. Of losing)
Suppose you draw a card from a deck of cards and are
paid $10 if it is an ace. Find the following expected
values.
If there is no cost:
Expectation = $10 x (4/52) ≈ $0.77
If the same game costs you $1 “up front” to play:
Expectation = $10 x (4/52) – $1 ≈ –$0.23
If the same game costs you $1 but you leave your money
“on the table” (if you win you get your $1 back).
Expectation = $10 x (4/52) – $1(48/52) ≈ –$0.15
Mathematical Expectation
If an event E has several possible outcomes with
probabilities p1, p2 , p3 , , pn , and if for each of these
outcomes the amount that can be won is
a1, a2 , a3 , , an , then the mathematical expectation
(or expected value) of E is:
Expectation  a1p1  a2 p2  a3 p3 
 an pn
Example
Suppose you play a game in which you have a
chance to win either $1, $5, $10, $20, or $100
depending on which bill you draw out of the box.
There are ten $1 bills, four $5 bills, three $10 bills,
two $20 bills, and one $100 bill. You pay $20 up front
to play the game. What is the expected value?
Should you play this game?
Expected value = $1(10/20) + $5(4/20) + $10(3/20)
+ $20(2/20) + $100(1/20) - $20
= -$10.00
You should NOT play this game!
Odds
Let
s = number of ways an event can occur
f = number of ways an event cannot occur
n = total number of possibilities
Then, the odds in favor of an event E = s/f
(the ratio of the number of ways the event can occur to the
number of ways it cannot occur)
And the odds against an event E = f/s
(the ratio of the number of ways the event cannot
occur to the number of ways it can occur)
Example
If a jar has 2 quarters, 200 dimes, and 800
pennies, and a coin is chosen at random,
what are the odds against picking a quarter?
Since there are 1000 ways of not picking a
quarter and there are only 2 ways of picking a
quarter, the odds against picking a quarter
(meaning the odds that you pick something
besides a quarter) = 1000/2 = 500/1
We say the odds against picking a quarter at
random are 500 to 1.
Finding the Odds Given the Probability
Let P(E) denote the probability of an event
occurring. Then the
P (E )
odds in favor of E =
P (E )
P (E )
odds against E =
P (E )
Finding the Probability Given the Odds in
Favor or the Odds Against an Event
If you know the odds in favor of an event
E (the ratio of s to f ), or if you know
the odds against an event E (the ratio
of f to s), then
s
P (E ) 
sf
and
f
P (E ) 
sf
If the probability of an event is 0.45, what
are the odds in favor of the event?
Since P(E) = 0.45 = 45/100 it follows that
the number of ways the event can occur is
45. This implies that the number of ways the
event cannot occur is 55 (because 45 + 55
= 100 which is the total number of
possibilities indicated by the value in the
denominator).
Thus, the odds in favor of E are 45 to 55
which reduces to 9 to 11.
If the odds against you are 20 to 1, what
is the probability of the event?
20 represents the number of ways the
event cannot occur.
1 represents the number of ways the event
can occur.
This implies that there are 21 total
possibilities, so the probability that the
event occurs = 1/21.