Download Probability Notes

Sec. 1.6
Probability
Objective: Students set up
probability equations
appropriately
Experimental
Probability
Probability
Number
of event =
of times event occurs
Number of trials
Example 1
A
player hit the bull’s eye on
a circular dartboard 8 times
out of 50. Find the
experimental probability that
the player hits the bull’s eye.
We need to use the
formula.
Number
of times event occurs =
Number of trials
8
 16%
50
Example 2
Find
the theoretical probability
of rolling a multiple of 3 with a
number cube? How about
rolling an odd?
The
Cube is a normal six
sided di.
 A)
How many numbers on the cube are
a multiple of 3?
 Yes 2 numbers, 3 and 6.
 So we get.
2 = 1
6
3
B) How many odds?
Yes 3 numbers, 1,3,5
So we get
3 = 1
6
2
Example 3
Suppose that all the points on the circular
dartboard shown below are equally likely
to be hit by a dart you have thrown. Find
the probability of only scoring 2 points
with one throw.
 Note: The radius of
each circle is one unit
larger than the one below
it.
20
20

10
5
2


First we need to find the area of the whole
dart board. This is the denominator because
any throw can hit any where on the dart
board.
To find the area of the green we need to
subtract the areas of the others. So we get
(using area πr2 of a circle)
π(4r)2 – π(3r)2
π(4r)2
= 16πr2 - 9πr2
16πr2
7
= 7πr2

2
16πr
16

20
20
10
5
2
Evaluate the expression
1) 3!
2)
7!
=3•2•1
=7•6•5•4•3•2•1
=6
= 5040
3) ( 4!/6! )
= (4 • 3 • 2 • 1) / (6 • 5 • 4 • 3 • 2 • 1)
= 1/(6 • 5)
= 1/30
Objective – To be able to use permutations to count the
number of ways an event can happen
State Standard –
18.0 Students use the fundamental counting
principles to compute combinations and
permutations.
19.0 Students use Combinations and Permutations
to compute probabilities.
The Fundamental Counting Principle
Suppose you own a small deli. You offer 4 types of meat (ham,
turkey, roast beef, and pastrami) and 3 types of bread (white,
wheat, and rye). How many choices do your customers have for a
meat sandwich?
White
Ham
Wheat
Rye
White
Wheat
Turkey
Rye
White
Roast Beef
Wheat
Rye
Pastrami
White
Pastrami on White
Wheat
Pastrami on Wheat
Rye
Pastrami on Rye
Example 1a
At a restaurant, you have a choice of 8 different entrees, 2 different
salads, 12 different drinks, and 6 different desserts. How many
different dinners consisting of 1 salad, 1 entrée, 1 drink, and 1 dessert
can you choose?
8 • 2 • 12 • 6 =
1152 different dinners
Example 1b
7th digit
6th digit
5th digit
4th digit
3rd digit
2nd digit
1st digit
How many different 7 digit phone numbers are possible if the first
digit cannot be 0 or 1?
8 • 10 • 10• 10 • 10 • 10 • 10
= 8,000,000
A Permutation – is an ordering of n objects.
For instance, there are 6 permutations of the letters A, B, and C:
ABC, ACB, BAC, BCA, CAB, CBA
In general there are 3 choices for the 1st letter, 2 choices for
the 2nd letter, and 1 choice for the 3rd letter so there are:
3 • 2 • 1 =6 ways to arrange the letters.
n!
Permutations of n objects taken r at a time
The number of permutations of r objects taken from a
group of n distinct objects is denoted by nPr and is
given by:
nPr = n!/(n – r)!
Example 2
Find the number of permutations.
5 P2
nPr = n!/(n – r)!
5 P2
= 5!/(5 – 2)!
=5•4•3•2•1
3•2•1
=5•4
= 20
Find the Product:
1) (2x – 3)2
= (2x – 3)(2x – 3)
2) (x – 3y)2
= (x – 3y)(x – 3y)
= 4x2 – 12x + 9
= x2 – 6xy + 9y2
Objective – To be able to use combinations to count the
number of ways an event can happen
State Standard – 19.0 Students use Combinations and
Permutations to compute probabilities.
Combinations of n objects taken r at a time
The number of combinations of r objects taken from a
group of n distinct objects is denoted by nCr and is
given by:
n!
nCr =
(n – r)! • r!
Example 3a
Find the number of combinations.
5 C2
nCr = n!/{(n – r)! • r!}
5 C2
= 5!/{(5 – 2)! • 2!}
= 5 • 4 • 3!
3! • 2!
=5•4
2
= 10
Example 3b
Find the number of combinations.
10 C4
nCr = n!/{(n – r)! • r!}
10 C4
= 10!/{(10 – 4)! • 4!}
= 10 • 9 • 8 • 7 • 6!
6! • 4!
3
= 10 • 9 • 8 • 7
4•3•2•1
= 210
Example 4
Use a standard deck of 52 cards.
a) If the order is not important, how many different
7-card hands are possible?
52!
52C7
=
(52 – 7)! • 7!
=133,784,560
b) How many of these hands have exactly 6 cards of
the same suit?
4C1 • 13C6 = 6,864
When finding the # of ways both an event A and an
event B can occur you need to multiply.
A bag contains 24 green marbles, 22 blue marbles, 14
yellow marbles, and 12 red marbles. Suppose you
pick one marble at random. Find each probability.
1) P(yellow)
14  7
36
2)
50
72
72
36
3) P(green or red)
72
P(not blue)
1

2
25

36
Objective – To be able to find the probabilities of events A
and B, and of events A or B.
State Standard –
1.0 Students will know the definition of the notion
of independent events and can use the rules of addition,
multiplication, and complementation to solve for
probabilities of particular events in finite sample spaces.
Dependent Events – Is when the outcome of one event
affects the outcome of a second event.
Independent Events – Is when the outcome of one event
does not affect the outcome of a second event.
Example 1
Classify each pair of events as dependent or independent.
a) Toss a coin. Then, select a marble from a bag that
contains marbles of different colors.
b) Select a marble from a bag that contains marbles of two
different colors. Put the marble aside, and select a second
marble from the bag.
Checking for Understanding:
Suppose you select a marble from a bag of marbles. You
replace the marble and then select again. Are your
selections dependent or independent event? Explain.
Independent; The number of marbles is the same after the
marble is replaced.
Probability of A and B
If A and B are independent events, the
probability that both A and B occur is:
P(A and B) = P(A)·P(B)
Example 2:
A random number generator on a computer selects
three integers from 1 to 20. What is the
probability that all three numbers are less than
or equal to 5?
The probability of selecting a number from 1 to
5 is:
1

5
P ( A) 
4
20
So the probability that all three numbers will
be less than or equal to 5 is:
 1  1  1 
1
P( A)  P( A)  P( A)     

 4  4  4 
64
Example 3:
A person draws 2 cards from a deck of 52
cards. What is the probability that the 2
cards drawn will both be face cards if the deck
contains 12 face cards?
12
P ( A) 
52
11
P( B) 
51
 12  11 
P( A)  P( B)    
 52  51 
132
11


= 0.05
2652
221
Probability
12.2
MATHPOWERTM 12, WESTERN EDITION 8.4.1
Conditional Probability
If A and B are events from an experiment, the conditional
probability of B given A (P(A|B)), is the probability that
Event B will occur given that Event A has already occurred.
The conditional probability is equal to the probability that
B and A will occur divided by the probability that B will occur.
This is given in Bayes’ Formula:
P(B and A)
P(A | B) 
, where P(B)  0
P(B)
8.4.2
Conditional Probability
Determine the conditional probability for each of the following:
a) Given P(B and A) = 0.725 and P(B) = 0.78, find P(A|B).
P(B and A)
P(A | B) 
P(B)
P(A|B) = 0.9295
0.725
P(A | B) 
0.78
a) Given P(blonde and tall) = 0.5 and P(blonde) = 0.73,
find P(A|B).
P(B and A)
P(A | B) 
P(B)
0.5
P(A | B) 
P(A|B) = 0.6849
0.73
8.4.3
Conditional Probability
LESSON 12-2
Additional Examples
The table shows the results of a class survey.
Find P(own a pet | female)
Do you own a pet?
yes
female
8
male
5
no
6
7
The condition female limits the sample space to 14 possible outcomes.
Of the 14 females, 8 own a pet.
Therefore, P(own a pet | female) equals 8 .
14
Conditional Probability
LESSON 12-2
Additional Examples
Using the data in the table, find the probability that a sample of
non-recycled waste was plastic.
The given condition limits the
sample space to non-recycled
waste.
A favorable outcome is
non-recycled plastic.
Material Recycled
Paper
36.7
Metal
6.3
Glass
2.4
Plastic
1.4
Other
21.2
Non-Recycled
45.1
11.9
10.1
24.0
70.1
24.0
45.1 + 11.9 + 10.1 + 24.0 + 70.1
24.0
=
161.2
0.15
The probability that the non-recycled waste was plastic is about 15%.
P(plastic | non-recycled) =
Conditional Probability
LESSON 12-2
Additional Examples
Researchers asked people who exercise regularly whether
they jog or walk. Fifty-eight percent of the respondents were male.
Twenty percent of all respondents were males who said they jog. Find
the probability that a male respondent jogs.
Relate: P( male ) = 58%
P( male and jogs ) = 20%
Define: Let A = male.
Let B = jogs.
Write:
P( A | B ) =
=
P( A and B )
P( A )
0.2
0.58
Substitute 0.2 for P(A and B) and 0.58 for P(A).
0.344
Simplify.
The probability that a male respondent jogs is about 34%.
Conditional Probability
LESSON 12-2
Additional Examples
A student made the following
observations of the weather in his
hometown.
• On 28% of the days, the sky was
mostly clear.
• During the mostly clear days, it rained
4% of the time.
P(rain|clear)
P(norain|clear)
P(rain|cloudy)
P(norain|cloudy)
• During the cloudy days it rained
31% of the time.
Use a tree diagram to find the probability that a day will
start out clear, and then it will rain.
Conditional Probability
LESSON 12-2
Additional Examples
(continued)
The path containing clear and rain represent days that start
out clear and then will rain.
P(clear and rain) = P(rain | clear) • P(clear)
= 0.04 • 0.28
= 0.011
The probability that a day will start out clear and then rain is
about 1%.
12.4 Warm Up
Find the mean, median and mode of the data set.
1) 6, 7, 9, 9, 9,10
Mean: x = 6+7+9+9+9+10 = 50 = 8.33
6
6
Median: 9
Mode: 9
Standard
Deviation
Standard Deviation
standard deviation- is often very important in scientific
experiments. A small standard deviation signifies the results
are consistent. A large standard deviation signifies the results
are inconclusive.
The standard deviation (read as “sigma”) of x1,x2…. xn
 = (x1 – x )2+(x2 – x )2+…+(xn - x)2
n
Example 1
What is the S.D. of the 3 test scores?
86, 89, and 91
Mean: x = 86 + 89 + 91 = 88.67
3
 = (x1 – x )2+(x2 – x )2+…+(xn - x)2
n
 = 2.06
Example 2
What is the S.D. of the 4 test scores?
70, 85, 99 and 40
Mean: x = 70 + 85 + 99 + 40 = 294 = 73.5
4
4
 = (x1 – x )2+(x2 – x )2+…+(xn - x)2
n
 = 21.89