Transcript Ch13-Sec13.3

13
THE NATURE
OF PROBABILITY
Copyright © Cengage Learning. All rights reserved.
13.3
Probability Models
Copyright © Cengage Learning. All rights reserved.
Complementary Probabilities
3
Complementary Probabilities
Let
s = NUMBER OF GOOD OUTCOMES (successes)
f = NUMBER OF BAD OUTCOMES (failures)
n = TOTAL NUMBER OF POSSIBLE OUTCOMES
(s + f = n)
4
Complementary Probabilities
Then the probability that event E occurs is
P(E) =
The probability that event E does not occur is
P( ) =
5
Complementary Probabilities
An important property is found by adding these
probabilities:
P(E) + P(
)=
=
=
=1
6
Complementary Probabilities
The probabilities P(E) and P( ) are called
complementary probabilities.
In other words, the property of complements says that
probabilities whose sum is 1 are complementary.
7
Example 2 – Find the probability of at least one head
What is the probability of obtaining at least one head in
three flips of a coin?
Solution:
Let F = {obtain at least one head in three flips of a coin}.
8
Example 2 – Solution
cont’d
Method I:
Work directly; use a tree diagram to find the possibilities.
P(F) =
9
Example 2 – Solution
cont’d
Method II:
For one coin, there are 2 outcomes (heads and tails); for
three coins we have
2  2  2 = 8 possibilities
Fundamental counting principle
We answer the question by finding the complement; is
the event of receiving no heads (that is, of obtaining all
tails).
10
Example 2 – Solution
cont’d
Without drawing the tree diagram, we note that there is
only one way of obtaining all tails (TTT).
Thus,
P(F) = 1 –
11
Odds
12
Odds
Related to probability is the notion of odds. Instead of
forming ratios
P(E) =
and
=
we form the following ratios:
Odds in favor of an event E:
Odds against an event E:
(ratio of success to failure)
(ratio of failure to success)
13
Odds
As we know that: s = NUMBER OF SUCCESSES
f = NUMBER OF FAILURES
n = NUMBER OF POSSIBILITIES
14
Example 3 – Find odds against
If a jar has 2 quarters, 200 dimes, and 800 pennies, and a
coin is to be chosen at random, what are the odds against
picking a quarter?
Solution:
We are interested in picking a quarter, so let s = 2; a failure
is not obtaining a quarter, so f = 1,000. (Find f by adding
200 and 800.)
Odds in favor of obtaining a quarter:
15
Example 3 – Solution
cont’d
Odds against obtaining a quarter:
=
=
Do not write
as 500.
The odds against picking a quarter when a coin is selected
at random are 500 to 1.
16
Odds
Sometimes you know the probability and want to find the
odds, or you may know the odds and want to find the
probability.
These relationships are easy if you remember:
s+f=n
17
Odds
You can show these formulas to be true:
18
Odds
19
Example 5 – Find the odds in favor
If the probability of an event is 0.45, what are the odds in
favor of the event?
Solution:
P(E) = 0.45
This is given
=
=
P( ) = 1 –
=
20
Example 5 – Solution
cont’d
Then the odds in favor of E are
The odds in favor are 9 to 11.
21
Example 6 – Find a probability given the odds against
If the odds against you are 20 to 1, what is the probability
of the event?
Solution:
Odds against an event are f to s, so f = 20 and s = 1;
thus,
P(E) =
 0.048
22
Example 7 – Find a probability, given the odds in favor
If the odds in favor of some event are 2 to 5, what is the
probability of the event?
Solution:
Odds in favor are s to f, so s = 2 and f = 5;
thus,
P(E) =
=
 0.286
23
Conditional Probability
24
Conditional Probability
Suppose that a family has two children. What is the
probability that the family has two boys?
Sample space: BB, BG, GB, GG;
1 success out of 4 possibilities
Now, let’s complicate the problem a little. Suppose that we
know that the older child is a boy.
25
Conditional Probability
We have altered the sample space as follows:
Original sample space: BB, BG, GB, GG; but we need to
cross out the last two possibilities because we know that
the older child is a boy:
Altered sample space:
26
Conditional Probability
P(2 boys given the older is a boy) =
This is a problem involving a conditional
probability—namely, a probability of an event given that
another event F has occurred. We denote this by
P(E|F)
Read this as: “probability of E given F.”
27
Conditional Probability
If the sample space consists of mutually exclusive and
equally likely possibilities, then P(E|F) represents the
number of outcomes corresponding to E ∩ F over the
number of outcomes corresponding to F.
That is, in “E given F” we are taking F as “given” by putting
only the outcomes in F into the denominator in the
definition of probability.
28
Conditional Probability
In the numerator, we have the outcomes of E that are also
in F. This means that
We call this the conditional probability formula. While this
formula is useful in probability theory, we will generally find
conditional probability by looking at altered sample spaces,
as illustrated by the next example.
29
Example 8 – Find a probability using an altered sample space
Suppose that you toss two coins (or a single coin twice).
What is the probability that two heads are obtained if you
know that at least one head is obtained?
Solution:
Consider an altered sample space: HH, HT, TH,
The probability is
30
Example 10 – Find the probability of drawing two cards
Two cards are drawn from a deck of cards. Find the
probability of the given events.
a. The first card drawn is a heart.
b. The first card drawn is not a heart.
c. The second card drawn is a heart if the first card drawn
was a heart.
d. The second card drawn is not a heart if the first card
drawn was a heart.
31
Example 10 – Find the probability of drawing two cards
cont’d
e. The second card drawn is a heart if the first card drawn
was not a heart.
f. The second card drawn is not a heart if the first card
drawn was not a heart.
g. The second card drawn is a heart.
h. Use a tree diagram to represent the indicated
probabilities.
32
Example 10 – Solution
Let H1 = {first card drawn is a heart},
H2 = {second card drawn is a heart}.
a.
There are 13 hearts in a deck of 52 cards.
b.
There are 39 nonhearts in a deck of 52 cards.
c.
If a heart is obtained on the first draw, then for
the second draw there are 12 hearts in the deck
of 51 remaining cards.
33
Example 10 – Solution
cont’d
d.
If a heart is obtained on the first draw, then
for the second draw there are 39 nonhearts
in a deck of 51 remaining cards.
e.
If a heart is not obtained on the first draw, then
for the second draw there are 13 hearts in a
deck of 51 remaining cards.
f.
If a nonheart is obtained on the first draw, then
for the second draw there are 38 nonhearts in a
deck of 51 cards.
g.
If we do not know what happened on the first
draw, then this is not a conditional probability, so
we consider this probability to be the same as
P(H1).
34
Example 10 – Solution
cont’d
h. We use a tree diagram to illustrate this situation.
35
Example 10 – Solution
cont’d
Note that we use the word limb to mean “a sequence of
branches that starts at the beginning.”
Also note that we have filled in the probabilities on each
branch.
To find the probabilities, we multiply as we move
horizontally across a limb, and we add as we move
vertically from limb to limb.
36
Example 10 – Solution
cont’d
Verify that the probabilities for the first card add to 1:
Also, for the second card, the probabilities for each branch
add to 1.
37
Example 10 – Solution
cont’d
For P(H2), we can confirm our reasoning in part g by
considering the first and third limbs of the tree diagram:
38
Example 10 – Solution
cont’d
Notice that conditional probabilities are found in the tree
diagram by beginning at their condition.
Verify that all of the information in parts a–g can be found
directly from the tree diagram.
For this reason, we often use tree diagrams to assist us in
finding probabilities and conditional probabilities.
39
Conditional Probability
40