Section 11.5

Download Report

Transcript Section 11.5

Chapter 11
Probability
Copyright © 2015, 2011, and 2007 Pearson Education, Inc.
1
Chapter 11: Probability
11.1
11.2
11.3
11.4
11.5
Basic Concepts
Events Involving “Not” and “Or”
Conditional Probability and Events
Involving “And”
Binomial Probability
Expected Value and Simulation
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
2
Section 11-5
Expected Value and Simulation
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
3
Expected Value
• Determine expected value of a random variable
and expected net winnings in a game of
chance.
• Determine if a game of chance is a fair game.
• Use expected value to make business and
insurance decisions.
• Use simulation in genetic processes such as
flower color and birth gender.
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
4
Expected Value
Children in third grade were surveyed and told to pick
the number of hours that they play electronic games
each day. The probability distribution is given below.
# of Hours x
Probability P(x)
0
0.3
1
0.4
2
0.2
3
0.1
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
5
Expected Value
Compute a “weighted average” by multiplying
each possible time value by its probability and
then adding the products.
0(0.3)  1(0.4)  2(0.2)  3(0.1)  1.1
1.1 hours is the expected value (or the
mathematical expectation) of the quantity of
time spent playing electronic games.
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
6
Expected Value
If a random variable x can have any of the values
x1, x2 , x3 ,…, xn, and the corresponding
probabilities of these values occurring are
P(x1), P(x2), P(x3), …, P(xn), then the expected
value of x is given by
E ( x)  x1  P( x1 )  x2  P( x2 ) 
 xn  P( xn ).
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
7
Example: Finding the Expected
Number of Boys
Find the expected number of boys for a three-child
family. Assume girls and boys are equally likely.
Solution
S = {ggg, ggb, gbg,
gbb, bgg, bgb, bbg,
bbb}
The probability
distribution is on
the right.
# Boys
x
0
1
2
3
Probability
P(x)
1/8
3/8
3/8
1/8
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
Product
x  P( x)
0
3/8
6/8
3/8
8
Example: Finding the Expected
Number of Boys
Solution (continued)
The expected value is the sum of the third column:
3 6 3 12
0   
8 8 8 8
3
  1.5.
2
So the expected number of boys is 1.5.
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
9
Example: Finding Expected Winnings
A player pays $3 to play the following game: He rolls
a die and receives $7 if he tosses a 6 and $1 for
anything else. Find the player’s expected net
winnings for the game.
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
10
Example: Finding Expected Winnings
Solution
The information for the game is displayed below.
Die Outcome
Payoff
1, 2, 3, 4, or 5
$1
Net P(x)
x
–$2 5/6
6
$7
$4
1/6
x  P( x)
–$10/6
$4/6
Expected value: E(x) = –$6/6 = –$1.00
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
11
Games and Gambling
A game in which the expected net winnings are
zero is called a fair game. A game with negative
expected winnings is unfair against the player. A
game with positive expected net winnings is
unfair in favor of the player.
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
12
Example: Finding the Cost for a Fair
Game
What should the game in the previous example
cost so that it is a fair game?
Solution
Because the cost of $3 resulted in a net loss of $1,
we can conclude that the $3 cost was $1 too high. A
fair cost to play the game would be $3 – $1 = $2.
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
13
Example: Finding Expected Winnings
in Roulette
One simple type of roulette is played with an ivory
ball and a wheel set in motion. The wheel contains
thirty-eight compartments. Eighteen of the
compartments are black, eighteen are red, one is
labeled “zero,” and one is labeled “double zero.”
(These last two are neither black nor red.) In this
case, assume the player places $1 on either red or
black. If the player picks the correct color of the
compartment in which the ball finally lands, the
payoff is $2. Otherwise, the payoff is zero. Find the
expected net winnings.
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
14
Example: Finding Expected Winnings
in Roulette
Solution
By the expected value formula, expected net
winnings are
18
20
1
E (net winnings)  ($1)  ($1)  $
38
38
19
The expected net loss here is $1/19, or about 5.3¢,
per play.
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
15
Investments
Expected value can be a useful tool for
evaluating investment opportunities.
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
16
Example: Finding Expected Investment
Profits
Mark is going to invest in the stock of one of the two
companies below. Based on his research, a $6000
investment could give the following returns.
Company ABC
Profit or Probability
Loss x
P(x)
–$400
0.2
$800
0.5
$1300
0.3
Company PDQ
Profit or
Probability
Loss x
P(x)
$600
0.8
1000
0.2
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
17
Example: Finding Expected Investment
Profits
Find the expected profit (or loss) for each of the
two stocks.
Solution
ABC: –$400(0.2) + $800(0.5) + $1300(0.3) = $710
PDQ: $600(0.8) + $1000(0.2) = $680
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
18
Business and Insurance
Expected value can be used to help make
decisions in various areas of business, including
insurance.
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
19
Example: Expected Lumber Revenue
A lumber wholesaler is planning on purchasing a
load of lumber. He calculates that the probabilities
of reselling the load for $9500, $9000, or $8500 are
0.25, 0.60, and 0.15, respectively. In order to ensure
an expected profit of at least $2500, how much can
he afford to pay for the load?
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
20
Example: Expected Lumber Revenue
Solution
The expected revenue from sales can be found below.
Income x
P(x)
x  P( x)
$9500
.25
$2375
$9000
.60
$5400
$8500
.15
$1275
Expected revenue: E(x) = $9050
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
21
Example: Expected Lumber Revenue
Solution (continued)
expected profit = expected revenue – cost
cost = expected revenue – expected profit
To have an expected profit of $2500, he can pay up to
$9050 – $2500 = $6550.
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
22
Simulation
An important area within probability theory is the
process called simulation. It is possible to study a
complicated, or unclear, phenomenon by simulating,
or imitating, it with a simpler phenomenon
involving the same basic probabilities. Simulation
methods (also called Monte Carlo methods)
require huge numbers of random digits, so
computers are used to produce them.
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
23
Example: Simulating Genetic
Processes
Toss two coins 50 times and use the results to
approximate the probability that the crossing of Rr
pea plants will produce three successive redflowered offspring.
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
24
Example: Simulating Genetic
Processes
Solution
We actually tossed two coins 50 times and got the
following sequence.
th, hh, th, tt, th, hh, ht, th, ht, th, hh, hh, tt, th, hh,
ht, ht, ht, ht, th, hh, hh, hh, tt, ht, tt, hh, ht, ht, hh, tt,
tt, tt, th, tt, tt, hh, ht, ht, ht, hh, tt, th, hh, tt, hh, ht,
tt, tt, tt
By the color interpretation described in the text, this
gives the following sequence of flower colors in the
offspring. (Only “both tails” gives white.)
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
25
Example: Simulating Genetic
Processes
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
26
Example: Simulating Genetic
Processes
We now have an experimental list of 48 sets of three
successive plants, the 1st, 2nd, and 3rd entries, then
the 2nd, 3rd, and 4th entries, and so on. Do you see
why there are 48 in all?
Now we just count up the number of these sets of
three that are “red-red-red.” Since there are 20 of
those, our empirical probability of three successive
red offspring, obtained through simulation, is 20/48
= 5/12, or about 0.417.
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
27
Example: Simulating Births with
Random Numbers
A couple plans to have five children. Use random
number simulation to estimate the probability that they
will have more than three boys.
Solution
Let each sequence of five digits, as they appear in Table
20 (page 632 from text), represent a family with five
children, and (arbitrarily) associate odd digits with boys,
even digits with girls. (Recall that 0 is even.) Verify
that, of the fifty families simulated, only the ten marked
with arrows have more than 3 boys (4 boys or 5 boys).
P(more than 3 boys)  10 / 50  0.20
Copyright © 2016, 2012, and 2008 Pearson Education, Inc.
28