Transcript P - WebAssign

13
THE NATURE
OF PROBABILITY
Copyright © Cengage Learning. All rights reserved.
13.4
Calculated Probabilities
Copyright © Cengage Learning. All rights reserved.
Keno Games
3
Keno Games
A very popular lotto game in several states, as well as in
most casinos, is a game called Keno.
The game consists of a player’s trying to guess in advance
which numbers will be selected from a pot containing 80
numbers.
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Keno Games
The person’s choices are marked on a Keno card, and
those choices are registered before the game starts.
Twenty numbers are then selected at random.
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Keno Games
The player may choose from 1 to 15 spots and gets paid
according to Table 13.2 on below.
Keno Payoffs
Table 13.2
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Keno Games
cont’d
Keno Payoffs
Table 13.2
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Example 1 – Decide if a game is fair
Suppose a person picks one number and is paid $3.00 if
the number picked is among the 20 chosen for the game.
The cost for playing this game (the cost is collected in
advance) is $1.00. Is this a fair game?
Solution:
In this problem we are picking 1 out of 80 and we will win if
the one number we pick is among the 20 numbers chosen
from the pot.
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Example 1 – Solution
cont’d
Since the order in which the 20 numbers are picked from
the pot does not matter, we see this is a combination.
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Example 1 – Solution
cont’d
We now calculate the mathematical expectation:
EXPECTATION = $3.00
– $1.00 = –$0.25
Subtract because you pay first.
No, it is not a fair game, since the expectation is negative.
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Independent Events
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Independent Events
Consider the following problem dealing with four cards.
Suppose four cards are taken from an ordinary deck of
cards, and we form two stacks—one with an ace (one) and
a deuce (two) and the other with an ace and a jack.
If a card is drawn at random from each pile, what is the
probability that a blackjack will occur (an ace and a jack)?
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Independent Events
The tree diagram illustrates the possibilities.
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Independent Events
Notice that the probability of obtaining an ace from the first
stack is 1/2 and the probability of obtaining a jack from the
second stack is also 1/2.
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Independent Events
The probability of a blackjack is . Notice for this example
that
probability of ace from first stack =
probability of jack from second stack =
probability of blackjack =
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Independent Events
If one event (draw from the first stack) has no effect on the
outcome of the second event (draw from the second stack),
then we say that the events are independent.
Thus, if events E and F are independent, P(E|F) = P(E),
since the occurrence of event F has no effect on the
occurrence of E.
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Independent Events
To understand how independence is determined, consider
the following alternative problem. Suppose that all four
cards are put together into one pile. This is an entirely
different situation.
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Independent Events
There are four possibilities for the
first draw, and three possibilities
for the second draw, as
illustrated here.
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Independent Events
The probability of a blackjack is now
.
We find this by looking at the sample space of equally likely
possibilities. We cannot find this probability by multiplying
the component parts for the first and second draws, as we
did when considering independent stacks.
For this second situation we see that the first and second
draws are dependent because the probabilities of
selecting a particular second card are influenced by the
draw on the first card.
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Probability of an Intersection
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Probability of an Intersection
If two events are independent, then we can find the
probability of an intersection by multiplication. This property
is called the multiplication property of probability or the
probability of an intersection.
If there are n mutually exclusive and equally likely
possibilities, then by definition
By the conditional probability formula
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Probability of an Intersection
= P(E|F)  P(F)
By definition of probability
= P(E)  P(F)
P(E|F) = P(E)
since E and F are independent.
This leads us to the following property.
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Example 3 – Find a roulette probability
What is the probability of black occurring on two successive
plays on a U.S. roulette wheel?
Solution:
A U.S. roulette wheel has 38 compartments, 18 of which
are black. Since the results of the spin of the wheel for one
game are independent of the results of another spin, the
plays are independent. Let B1 = {black on first spin} and
B2 = {black on second spin}.
Then
 0.2243767313
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Probability of an Intersection
Independence and the multiplication property are often
used with the idea of complementary events.
We now consider the mathematics associated with an
experiment called the birthday problem.
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Example 4 – Birthday problem
What is the probability that at least 2 of 4 unrelated persons
share the same birthday?
Solution:
We use Pólya’s problem-solving guidelines for this
example.
Understand the Problem. We assume that the birthdays
of the individuals are independent, and we ignore leap
years. We also are not considering the birthday year. We
want to know, for example, whether any 2 (or more) of the
4 persons have the same birthday—say, August 3.
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Example 4 – Solution
cont’d
Devise a Plan. The first person’s birthday can be any day
(365 possibilities out of 365 days). We will find the
probability that the second person’s birthday is different
from the first person’s birthday. This is
The probability that the third person’s birthday is different
from the first two birthdays is
, and the probability that
the fourth person’s birthday is different is
. Since these
are independent events, we use the multiplication property
of probability.
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Example 4 – Solution
cont’d
Carry Out the Plan.
P(match) = 1 – P(no match)
 0.0163559125
=
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Example 4 – Solution
cont’d
Look Back. There are about 2 chances out of 100 that
there will be a birthday match with 4 persons in the group.
It is worth noting that if we consider this for 23 persons, we
have
This means that if you look at a group of more than 23
persons, it is more likely than not that there will be a
birthday match.
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Probability of a Union
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Probability of a Union
The multiplication property of probability is used to find the
probability of an intersection (E and F); now let’s turn to the
probability of a union (E or F).
P(E or F) = P(E  F)
Translate the word
“or” as a union.
=
Definition of probability
=
Cardinality of a union
= P(E) + P(F) – P(E  F)
Definition of probability
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Probability of a Union
We have derived a property called the addition property
of probability.
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Example 6 – Find a calculated probability
If A, B, and C are independent events so that P(A) = 0.2,
P(B) = 0.6, and P(C) = 0.3, find P[(A  B)  C].
Solution:
We need to do some algebra before we evaluate.
P[(A  B)  C] = 1 – P[(A  B)  C]
= 1 – P[(A  B)  C]
Complementary probabilities
Multiplication property
of probability
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Example 6 – Solution
cont’d
= 1 – [P(A) + P(B) – P(A  B)]  P(C)
Addition property
= 1 – [P(A) + P(B) – P(A)  P(B)]  P(C)
Multiplication property
= 1 – P(A)  P(C) – P(B)  P(C) + P(A)  P(B)  P(C)
Distributive property (watch signs)
We now evaluate:
P[(A  B)  C] = 1 – (0.2)(0.3) – (0.6)(0.3) + (0.2)(0.6)(0.3)
= 0.796
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Drawing With and Without Replacement
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Drawing With and Without Replacement
To highlight the difference between a combination model
and a permutation model, we focus on the idea of
replacement.
Drawing with replacement means choosing the first item,
noting the result, and then replacing the item in the sample
space before selecting the second item.
Drawing without replacement means selecting the first
item, noting the result, and then selecting a second item
without replacing the first item.
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Example 7 – Find probabilities with and without replacement
Find the probability of each event when drawing two cards
from an ordinary deck of cards.
a. Drawing a spade on the first draw and a heart on the
second draw with replacement
b. Drawing a spade on the first draw or a heart on the
second draw with replacement
c. Drawing two hearts with replacement
d. Drawing a spade on the first draw and a heart on the
second draw without replacement
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Example 7 – Find probabilities with and without replacement
e. Drawing a spade on the first draw or a heart on the
second draw without replacement
f. Drawing two hearts without replacement
Rank the events from the most probable to the least
probable.
Solution:
We state the decimal approximations for each probability
for ease of comparison.
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Example 7 – Solution
cont’d
Let S1 = {draw a spade on the first draw} and
H2 = {draw a heart on the second draw}
With replacement
a. With replacement, the events are independent. Thus
P(S1  H2) = P(S1)  P(H2) =

=
= 0.0625
b. P(S1  H2) = P(S1) + P(H2) – P(S1  H2)
=
= 0.4375
From part a
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Example 7 – Solution
cont’d
c. With replacement, the events are independent. Then
P(two hearts) =
=

= 0.0625
Without replacement
d. We use the permutation model since the order is
important. The number of possibilities is 52P2 = 52  51;
the number of successes (fundamental counting
principle) is 13  13. Thus,
P(S1  H2) =
=
=
 0.0637254902
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Example 7 – Solution
cont’d
e. P(S1  H2) = P(S1) + P(H2) – P(S1  H2)
From part a
=
 0.4362745098
f. Without replacement, draw both cards at once. We use
combinations since the order is not important.
P(two hearts) =
=
=
 0.0588235294
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Example 7 – Solution
cont’d
The ranking is: (1) part b, (2) part e (probabilities of parts b
and e are about the same), (3) part d, (4) parts a and c
(tie), (5) part f. The probabilities of parts a, c, d, and f are
about the same.
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Tree Diagrams
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Tree Diagrams
A powerful tool in handling probability problems is a device
we have frequently used—a tree diagram.
If the events are independent, we find the probabilities
using the multiplication property of probability.
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Example 8 – Deciding to play. Or not?
Consider a game consisting of at most three cuts with a
deck of cards. You win and the game is over if a face card
turns up on any of the cuts, but you lose if a face card does
not turn up. (A face card is a jack, queen, or king.)
If you stand to win or lose $1 on this game, should you
play?
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Example 8 – Solution
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Example 8 – Solution
cont’d
P(win) = 1 – P(lose)  1 – 0.455 = 0.545
EXPECTATION = Amount to win  P(win)
+ Amount to lose  P(lose)
 $1  (0.545) + (–$1)  (0.455)
= $0.09
Since the mathematical expectation is positive, the
recommendation is to play the game. If you played this
game 1,000 times, you could expect to be ahead about
1,000($0.09) = $90.00
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