Download SIA Unit 4 Probability

UNIT 4
PROBABILITY
SCIENTIFIC INQUIRY AND ANALYSIS
1
OBJECTIVES UNIT 4
The student will be able to:
• Describe events as subsets of a set of outcomes using
characteristics or categories of the outcomes, or as
unions, intersections, or complements of other events
(“or”, “and”, “not”.) (CCSS.HSS.CP.A.1)
• Find the number of ways a group of objects can be
arranged in order (permutations with and without
replacement). (CCSS.HSS.CP.B.9)
• Find the number of ways to choose several objects from
a group without regard to order (combinations with and
without replacement). (CCSS.HSS.CP.B.9)
SCIENTIFIC INQUIRY AND ANALYSIS
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OBJECTIVES UNIT 4
• Find the probability of simple and compound events.
(5.3.12.C1)
• Determine if two events are independent by showing P(A
given B) is the same as P(A) and that the P(B given A) is
the same as P(B). (CCSS.HSS.CP.A.2, CCSS.HSS.CP.A.3)
• Find conditional probabilities as probability of A given B as
P(A and B)/P(B). (CCSS.HSS.CP.B.6)
• Use the multiplication rule to find the probability of two
events occurring in a sequence and to find conditional
probabilities. (CCSS.HSS.CP. B.8)
• Use the addition rule to find the probability of two events.
(CCSS.HSS.CP.B.7)
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Probability – the numerical measure of the
likelihood of an event occurring.
• Statistics – collection of methods for planning
experiments, obtaining data, and then
organizing, summarizing, presenting,
analyzing, interpreting, and drawing
conclusions from the data.
SCIENTIFIC INQUIRY AND ANALYSIS
4
PROBABILITY
• Sample Space
– Before analyzing the probability that something
can occur, it is necessary to understand the number
of possible outcomes given a set of possible
choices or selections. This is known as the sample
space.
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Fundamental Counting Principle
– When determining the number of possible
outcomes of an event, it is always not practical to
count all of the outcomes. All of the possible
outcomes is the sample space.
• One method to use to find the sample space is to use the
Tree – Method.
• Another method to use to find the sample space is to us
the Grid – Method. The grid – method is best if the
number of selections is no more than 2.
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Sample Space: (example using tree method)
If you have a pair of blue pants, a pair of black pants, a white
shirt and a red shirt, how many different ways can you wear one
of your pants and one of your shirts? What are the events in this
example?
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Sample Space: (example using grid method)
If you have a pair of blue pants, a pair of black pants, a
white shirt and a blue shirt, how many different ways can
you wear one of your pants and one of your shirts?
White Shirt
Blue Shirt
Blue Pants
White Shirt & Blue
Pants
Blue Shirt & Blue
Pants
Black Pants
White Shirt & Black
Pants
Blue Shirt & Black
Pants
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Counting Example 1:
– Kevin goes to a sandwich shop to order a
sandwich. There is a choice of 5 different breads,
8 different meats, 5 different cheeses and 3
different condiments.
• If Kevin chooses one of each, determine how many
different sandwiches can be made through a tree
diagram?
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
AMERICAN CHEESE
• Counting Example 1:
MUSTARD
MAYO
OLIVE OIL
SWISS CHEESE
PROVOLONE
CHEDDAR CHEESE
BOILED HAM
MUENSTER CHEESE
HONEY HAM
SALAMI
PASTRAMI
ROAST BEEF
TURKEY
CHICKEN
WHOLE WHEAT
BREAD
BOLAGNA
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Fundamental Counting Principle (Product
Rule)
– Instead of counting out every different occurrence
through a tree diagram, the fundamental counting
principle is used to determine the number of
occurrences.
– Principle: if one event can occur in m ways and
another event can occur in n ways, then the
number of ways that both events can occur is equal
to m · n. (or m · n · o · p · q · r . . .)
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Counting Example 1:
– So, if we revisit this example, use the counting
principle to answer the question.
– Jonah goes to a sandwich shop to order a
sandwich. There is a choice of 5 different breads,
8 different meats, 5 different cheeses and 3
different condiments.
• If Jonah chooses one of each, how many different
sandwiches can be made?
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Fundamental Counting Principle without
Repetition
– Example: Conner goes to a sandwich shop the next
day to order a sandwich. Again, there is a choice
of 5 different breads, 8 different meats, 5 different
cheeses and 3 different condiments.
• Conner is feeling hungrier than Jonah. If Conner
chooses three different meats and one each of
everything else, how many different sandwiches can be
made? So, for this problem, you can’t repeat the type of
meat.
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
AMERICAN CHEESE
• Fundamental Counting
Principle without Repetition
WHOLE WHEAT
BREAD
BOILED HAM
HONEY HAM
SALAMI
HONEY HAM
SALAMI
PASTRAMI
SALAMI
PASTRAMI
ROAST BEEF
PASTRAMI
ROAST BEEF
TURKEY
ROAST BEEF
TURKEY
CHICKEN
TURKEY
CHICKEN
BOLAGNA
CHICKEN
BOLAGNA
MUSTARD
MAYO
OLIVE OIL
SWISS CHEESE
PROVOLONE
CHEDDAR CHEESE
MUENSTER CHEESE
BOLAGNA
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Fundamental
Repetition
Counting
Principle
without
– If one event (choosing the first meat) can occur m
ways, if that same event is done a second time
(choosing the second meat) and if you don’t want the
same outcome (the same meat twice, no repetitions),
then the second event can occur m – 1 ways. Likewise,
if the same event is done a third time (choosing a third
meat) and if you don’t want the same outcome (no
repetitions of the meat), then the third event can occur
m – 2 ways.
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Fundamental
Repetition
Counting
Principle
without
– So, if we revisit this example, use the counting
principle without repetitions to answer the question.
– Connor goes to a sandwich shop the next day to order
a sandwich. Again, there is a choice of 5 different
breads, 8 different meats, 5 different cheeses and 3
different condiments.
• Conner is feeling hungrier than Jonah. If Conner chooses
three different meats and one each of everything else, how
many different sandwiches can be made? So, for this
problem, you can’t repeat the type of meat.
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Fundamental
Repetition
Counting
Principle
with
– Assuming the same example, Mike goes to a
sandwich shop to order a sandwich and the choices
are the same.
• Again Mike is feeling hungrier than Jonah. If Mike
chooses three meats, but for this example he can choose
each meat more than once, how many different
sandwiches can be made?
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Counting Principle Example
– In a high school, there are 273 freshmen, 291
sophomores, 252 juniors and 237 seniors. In how
many different ways can a committee of 1
freshman, 1 sophomore, 1 junior and 1 senior be
chosen?
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Fundamental Counting Principle (Sum
Rule)
– Instead of counting out every different occurrence
the fundamental counting principle is used to
determine the number of occurrences.
– Principle: if one event can be chosen from two
disjoint groups, then the number of ways that the
event can occur is equal to m + n. (or m + n + o
+ p + q + r . . .)
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Fundamental Counting Principle (Sum
Rule)
– Example: Chris goes into a restaurant and can
choose between 4 different chicken entrees or 3
different beef entrees or 5 different pasta entrees or
5 different seafood entrees. If he has one entrée,
how many different entrees can he have? If he has
two entrees, how many different ways can he
order?
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Factorials
– Factorial (m!) is a mathematical operation that
multiplies a number m by this method :
𝑚 × 𝑚 − 1 × (𝑚 − 2) ⋯ × 1
– Example:
5! = 5 × 4 × 3 × 2 × 1 = 120
– Note: 1! = 1 & 0! = 1
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Permutations – determines the number of ways n
elements (objects) can be arranged in order r
ways.
• Permutation is calculated by figuring out the
factorial.
𝑛!
𝑃 𝑛, 𝑟 = 𝑛𝑃𝑟 =
𝑛 − 𝑟 !
• P(n,r) is sometimes phrased: “Permutations of n
taken r times”
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Permutation properties
𝑊ℎ𝑒𝑛 𝑛 = 𝑟, 𝑡ℎ𝑒𝑛 𝑃 𝑛, 𝑟 = 𝑛𝑃𝑟 = 𝑛!
𝑊ℎ𝑒𝑛 𝑟 = 0, 𝑡ℎ𝑒𝑛 𝑃 𝑛, 0 = 𝑛𝑃0 = 1
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Permutation Example 1:
– You are manager of a baseball team. You have 9
players on the team who can bat anywhere in the
batting order. (i.e. 1st, 2nd, 3rd, 4th . . . 9th). How
many different batting orders could you have?
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Permutation Example 2:
– You have 9 players that can bat in the first 3
positions of the batting order. How many different
ways can the manager arrange the first 3 positions
of the batting order?
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Permutation Example 3:
– Next year you are taking Calculus I, Intro to
Engineering, Spanish II, Phys Ed, Physics I,
American Literature, US History, Chemistry.
Assume each class is offered during each of the 8
periods of the day. In how many different orders
can you schedule your classes?
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Permutations with Repetition: for the permutations taken
so far, all of the objects were distinct. If they are not
distinct, then the permutations are not indistinguishable.
• For example: How many orders of three are there of the
letters M – O – M? Which means how many permutations
of the 3 letters take 3 ways:
3!
3! 3 × 2 × 1
𝑃 3,3 = 3𝑃3 =
= =
=6
3 − 3 ! 0!
1
• However some of the outcomes are not indistinguishable.
MOM, OMM, MMO, MOM OMM, MMO
• In this case only 3 outcomes are distinguishable.
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Permutations with Repetition: the number of
distinguishable permutations of n objects
where one object is repeated q1 times, another
is repeated q2 times, etc.
𝑛!
𝑞1 ! × 𝑞2 ! × ⋯ × 𝑞𝑘 !
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Permutation with Repetition Example 1:
– Find the number of distinguishable permutations of
the letters in REBECCA.
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• When not concerned with the order of an
outcome, it is known at a combination.
• Examples:
– Being dealt 5 cards in a card game. Not concerned
with the order.
– Ordering an omelet with 1 cheese, 1 meat and 1
vegetable.
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Combinations – how many different r objects
out of n objects, no order.
𝑛!
𝐶 𝑛, 𝑟 = 𝑛𝐶𝑟 =
𝑛 − 𝑟 ! × 𝑟!
• Given the same number of n objects and r
objects, which is larger the combination or the
permutation?
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Combination Example 1:
– You have 9 players that can bat in the first 3
positions of the batting order. How many different
combinations of players can the manager utilize to
fill in the first 3 positions of the batting order?
– Remember: it is not important that the 3 players
selected to bat in the 3 positions bat 1st, 2nd or 3rd.
Compare this to permutation example 2.
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Combination Example 2:
– You are dealt 4 cards out of regular deck of 52.
How many different combinations of the 4 cards
out of the 52 can you get?
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Multiple Combination Events
– Sometimes there are multiple combination events
which require you to calculate multiple times and then
either multiply them or add them together.
– For example you are dealt 7 cards out of 52. How
many possible ways are there of getting 7 cards all of
the same suit? OR
– There are 5 field day events and you have to
participate in at least 2 events. How many different
combination of events can you participate?
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Multiple Combination Events (Multiplication)
– When finding the number of ways both one event
(Event A) and another event (Event B) can occur, take
the combination of both and multiply the two together.
– Example: you are dealt 7 cards out of 52. How many
are all of the same suit?
Event A (chose 1 out 4 suits)
Event B (chose 7 out of 13 cards in a suit)
4𝐶1 × 13𝐶7
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• DO NOW Problem
– Have out your homework.
– Do the following problem:
You are packing for a vacation. At home you have 10
shirts and 7 pairs of shorts.
1.
2.
3.
In how many different ways can you choose 4 pairs of
shorts to take on vacation?
In how many different ways can you choose 2 shirts to
wear on the 1st and 2nd days of vacation?
If you bring 4 pairs of shorts and 6 shirts, how many
different outfits can you make?
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Probability – a number between 0 – 1 which
indicates the likelihood the event will occur.
– Experiment – a process by which an outcome is
obtained.
– Sample Space (S) – a set that is composed of a
finite number of possible outcomes.
– Event (E) – any subset of the sample space.
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Probability
– Probability of an outcome in E will occur is the
ratio of the number of outcomes in E to the number
of outcomes in Sample Space (S).
OUTCOMES
EVENT (E)
SAMPLE SPACE (S)
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Probability
OUTCOMES
EVENT (A)
SAMPLE SPACE (S)
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑖𝑛 𝐴
𝑃 𝐴 =
𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Example
– Experiment – draw two cards out of a deck of 52
cards. What is the probability both will be a red
face card?
– Sample Space (S) = # of ways to select two cards
out of deck of 52.
– Event (E) = # of ways to get two red face cards.
– Probability (P) =
# 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑔𝑒𝑡 𝑎 𝑟𝑒𝑑 𝑓𝑎𝑐𝑒 𝑐𝑎𝑟𝑑
# 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑠𝑒𝑙𝑒𝑐𝑡 2 𝑐𝑎𝑟𝑑𝑠 𝑜𝑢𝑡 𝑜𝑓 52
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Example
using a
Punnett
Square
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Example of Punnett Square
– Experiment – genes that determine eye color are
either Aa or Bb. Each parent contributes an A or a
gene and a B or b gene. What is the probability
that a baby whose parents are both AaBb will have
green eyes? (3 lower case, 1 upper case)
– Sample Space (S) =
– Event (E) =
– Probability (P) =
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Combination/Probability
Example:
– The graph shows the % of 100
pizzas sold according to type. If
3 pizza were randomly chosen
out of the 100, what is the
probability that all 3 were thin
crusts?
– Sample Space (S) =
combinations of 3 pizzas out of
all 100 pizzas
– Event (E) = combinations of 3
thin crust pizza
– Probability (P) =
Stuffed
Crust
Thick Crust 2%
Pizzas Sold
23%
SCIENTIFIC INQUIRY AND ANALYSIS
Pan
Pizza
22%
Thin Crust
53%
43
PROBABILITY
• Counting Principle Probability Example:
– GP Clayton goes to a sandwich shop to order a
sandwich. There is a choice of 5 different breads,
8 different meats, 5 different cheeses and 3
different condiments.
• If GP Clayton chooses one of each, how many different
sandwiches can be made?
# of sandwiches = 5 x 8 x 5 x 3 = 600
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Counting Principle Probability Example
(Continued):
• Orest ran out of turkey. He is wondering what is the
probability that GP Clayton will ask for a sandwich with
turkey?
– Sample Space (S) = the number of sandwiches
that can be made.
– Event (E) = the number of sandwiches with turkey
– Probability (P) =
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Counting Principle Probability Example
(Continued):
• Orest ran out of salami and Swiss cheese. He is
wondering what is the probability that Kevin will ask
for a sandwich with salami or Swiss?
– Sample Space (S) = the number of sandwiches
that can be made.
– Event (E) = the number of sandwiches with salami
or the number of sandwiches with Swiss
– Probability (P) =
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Geometric Probability:
– Using a particular geometric shape to represent the
event space and the whole geometric shape
represents the sample space.
– Example: given the blue square is 10’ x 10’ and
the orange square is 7’ x 7’, what is the
probability that a dart will fall in the orange
square.
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Empirical Probability - the ratio of the
number of favorable outcomes to the total
number of trials.
• Law of Large Numbers
– The theorem that describes the result of
performing the same experiment a large number of
times. According to the law, the average of the
results obtained from a large number of trials
should be close to the expected value, and will
tend to become closer as more trials are performed.
SCIENTIFIC INQUIRY AND ANALYSIS
48
PROBABILITY
• Example of Telephone Numbers
– Experiment – What is the probability that given a
telephone book page that a telephone number ends
in 1?, 2?, 3?, 4?, 5?, 6?, 7?, 8?, 9?, 0?
– Theoretical Probability (P) =
– Sample Space (S) =
SCIENTIFIC INQUIRY AND ANALYSIS
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PROBABILITY
• Example of Telephone Numbers
EVENT
NUMBER OF
OUTCOMES
FREQUENCY = # OF
OUTCOMES/SAMPLE SPACE
Telephone # ends in 1
Telephone # ends in 2
Telephone # ends in 3
Telephone # ends in 4
Telephone # ends in 5
Telephone # ends in 6
Telephone # ends in 7
Telephone # ends in 8
Telephone # ends in 9
Telephone # ends in 0
SCIENTIFIC INQUIRY AND ANALYSIS
50
PROBABILITY
• DO NOW
– As part of a monthly inspection at a hospital, the
inspection team randomly selects reports from 8 of
the 84 nurses who are on duty. What is the
probability that none of the reports selected will be
from the 10 most experienced nurses on duty? OR
What is the probability that all of the reports
selected will be from the 74 least experienced
nurses on duty?
SCIENTIFIC INQUIRY AND ANALYSIS
51
PROBABILITY
• Unions & Intersections Review
– A SET is a collection of elements.
• For example Set A can be made up of the numbers {2, 3,
6, 8} Set B can be made up of the numbers {2, 6, 9, 12}
– A UNION of sets A or B means that the elements
belong to either SETS A or B.
– An INTERSECTION of sets A and B means that
the elements belong to SETS A and B
∪ 𝑚𝑒𝑎𝑛𝑠 𝑡ℎ𝑒 𝑢𝑛𝑖𝑜𝑛
∩ 𝑚𝑒𝑎𝑛𝑠 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛
𝐼𝑓 𝐴 ∩ 𝐵 = 𝑒𝑚𝑝𝑡𝑦 𝑠𝑒𝑡, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒𝑦 𝑎𝑟𝑒 𝑚𝑢𝑡𝑢𝑎𝑙𝑙𝑦 𝑒𝑥𝑐𝑙𝑢𝑠𝑖𝑣𝑒
SCIENTIFIC INQUIRY AND ANALYSIS
52
PROBABILITY
• Unions & Intersections Review
𝐴 ∪ 𝐵 𝑚𝑒𝑎𝑛𝑠 𝑡ℎ𝑒 𝑢𝑛𝑖𝑜𝑛 𝑜𝑓 𝑠𝑒𝑡𝑠 𝐴 𝑜𝑟 𝐵
𝐴 ∩ 𝐵 𝑚𝑒𝑎𝑛𝑠 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑒𝑡𝑠 𝐴 𝑎𝑛𝑑 𝐵
𝐼𝑓 𝐴 ∩ 𝐵 = 𝑒𝑚𝑝𝑡𝑦 𝑠𝑒𝑡, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒𝑦 𝑎𝑟𝑒 𝑚𝑢𝑡𝑢𝑎𝑙𝑙𝑦 𝑒𝑥𝑐𝑙𝑢𝑠𝑖𝑣𝑒
SCIENTIFIC INQUIRY AND ANALYSIS
53
PROBABILITY
• Probabilities of Unions & Intersections
– If A and B are events in the same sample space, then the
probability of A or B occurring is:
P(A U B) = P(A) + P(B) – P(A ∩ B)
P(A U B) can also be written P(A or B)
P(A ∩ B) can also be written P(A and B)
SCIENTIFIC INQUIRY AND ANALYSIS
54
PROBABILITY
• Probabilities of Unions & Intersections
– If A and B are mutually exclusive events in the same
sample space, then the probability of A or B occurring is:
P(A U B) = P(A) + P(B)
SCIENTIFIC INQUIRY AND ANALYSIS
55
PROBABILITY
• Union Example 1:
– One card is selected from a standard deck of 52
playing cards. What is the probability that the card
is a heart or a face card?
P(A or B) = P(A) + P(B) – P(A and B)
SCIENTIFIC INQUIRY AND ANALYSIS
56
PROBABILITY
• Union Example 2:
– One card is selected from a standard deck of 52
playing cards. What is the probability that the card
is a heart or a diamond card?
P(A or B) = P(A) + P(B) – P(A and B)
SCIENTIFIC INQUIRY AND ANALYSIS
57
PROBABILITY
• Union Example 3:
– In a poll of HS juniors, 6 out of 15 took a French
class and 11 out 15 took a math class. Fourteen
out of 15 students took French or Math. What is
the probability that a student took both French and
Math?
P(A or B) =
P(A) =
P(B) =
P(A and B) =
SCIENTIFIC INQUIRY AND ANALYSIS
58
PROBABILITY
• Complements
A’
A
– If A is an event in a sample space X, then the event
consisting of all outcomes in X that are not in A is
called the complement, A’:
P(A’) = 1 - P(A)
SCIENTIFIC INQUIRY AND ANALYSIS
59
PROBABILITY
• Complement Example 1:
– In throwing two 6 sided dice what is the
probability that the sum is not 11?
𝑷 𝒔𝒖𝒎 𝒏𝒐𝒕 𝟏𝟏 = 𝟏 − 𝑷(𝒔𝒖𝒎 𝒊𝒔 𝟏𝟏)
SCIENTIFIC INQUIRY AND ANALYSIS
60
PROBABILITY
• Example 2:
– The freezer case at a grocery store contains several
frozen pies: 10 apple, 4 peach, 6 blueberry, 5
pumpkin, 3 peanut butter, and some pecan. You
are shopping in a hurry and pick one of the pies
from the freezer without looking at the type. The
probability that the pie you picked is apple or
pecan is 13/22. How many pecan pies are in the
freezer case?
SCIENTIFIC INQUIRY AND ANALYSIS
61
PROBABILITY
• DO NOW
– A deck of UNO cards is made up of 108 cards.
(Twenty five each of red, yellow, blue and green,
and 8 are wild cards). Each player is randomly
dealt a 7 card hand.
• What is the probability that a hand will have 2 wild
cards?
• What is the probability that a hand will have 2 wild
cards, 2 red cards and 3 blue cards?
• What is the probability that a hand will have at least 2
wild cards?
SCIENTIFIC INQUIRY AND ANALYSIS
62
PROBABILITY
• Independent Events
– Two events are independent if the occurrence of
one has no effect on the occurrence of the other.
– Given A & B are independent events, the
probability that A & B occur is:
P(A ∩ B) = P(A) · P(B)
Remember that P(A ∩ B) can be written P(A and B)
SCIENTIFIC INQUIRY AND ANALYSIS
63
PROBABILITY
• Independent Events
– When calculating probability of independent
events, the probability can be found on more than
2 independent events.
– Given A & B & C are independent events, the
probability that A & B & C occur is:
P(A ∩ B ∩ C) = P(A) · P(B) · P(C)
– Typically for independent events the sample space stays the
same.
SCIENTIFIC INQUIRY AND ANALYSIS
64
PROBABILITY
• Independent Event Example 1:
– In a survey, 9 out of 11 men and 4 out of 7 women
said they were satisfied with a product. If the next
3 customers are 2 women and a man, what is the
probability that they will all be satisfied?
SCIENTIFIC INQUIRY AND ANALYSIS
65
PROBABILITY
• Independent Event Example 2:
– The following examples are better done using
complements.
– Given the previous survey of example 1, what is
the probability that if men are the next 4 customers
that at least 1 of them is not satisfied with the
product?
– Given the previous survey of example 1, what is
the probability that if men are the next 4 customers
that all of them are not satisfied with the product?
SCIENTIFIC INQUIRY AND ANALYSIS
66
PROBABILITY
• Dependent Events
– Two events are dependent if the occurrence of one
has an effect on the occurrence of the other.
– Given A & B are dependent events, the probability
that A & B occur is:
P(A ∩ B) = P(A) · P(B|A)
P(B|A) is known as a conditional probability that indicates
the probability that B occurs given that A has occurred.
SCIENTIFIC INQUIRY AND ANALYSIS
67
PROBABILITY
• Dependent Events
– When calculating probability of dependent events,
the probability can be dependent on more than 2
events.
– Given A & B & C are dependent events, the
probability that A & B & C occur is:
P(A ∩ B ∩ C) = P(A) · P(B|A) · P(C|A & B)
SCIENTIFIC INQUIRY AND ANALYSIS
68
PROBABILITY
• Dependent Event Example 1:
– Three children have a choice of 12 summer camps
that they can attend. If they each randomly choose
which camp to attend what is the probability that
they all attended different camps?
SCIENTIFIC INQUIRY AND ANALYSIS
69
PROBABILITY
• Independent vs. Dependent Events
– A box contains the numbers 1 – 20. What is the
probability of selecting 1 number less than 6,
replacing the number and selecting a second
number less than 6?
– Is this process dependent or independent?
– How can the above event be changed to the
opposite type of event?
SCIENTIFIC INQUIRY AND ANALYSIS
70
PROBABILITY
• Example:
– There are 25 pieces of paper, numbered 1 to 25, in
a hat. You pick a piece of paper, replace it and
then pick another piece of paper. What is the
probability that each number is greater than 20 or
less than 4?
SCIENTIFIC INQUIRY AND ANALYSIS
71
PROBABILITY
• Complements (Review)
– When finding the probability of something
occurring, it is sometimes easier to find the
probability when something won’t occur (its
complement) and then use the formula below.
P(A’) = 1 - P(A)
SCIENTIFIC INQUIRY AND ANALYSIS
72
PROBABILITY
• Complement Example:
– Suppose that 1% of the lights in a shipment of 10,000
lights are defective. If you make 100 random tests of
the lights (each time choosing, testing and replacing
the light), what is the probability that at least one of
the tests reveals a defective light.
– P(A’) = 9900 / 10,000 = 0.99 (selecting 1 good light)
– P(A’)100 = 0.99100 ≈ 0.366 (selecting 100 good lights,
100 times)
– P(A) = 1 - 0.99100 ≈ 0.634 (at least 1 bad after
selecting 100)
SCIENTIFIC INQUIRY AND ANALYSIS
73
PROBABILITY
• DO NOW:
In one town 95% of the students graduate from high
school. Suppose a study showed that at age 25, 81%
of the high school graduates held full-time jobs
while only 63% of those who did not graduate held
full-time jobs. What is the probability that a
randomly selected student from the town will have a
full time job at age 25?
SCIENTIFIC INQUIRY AND ANALYSIS
74
HOMEWORK UNIT 4-1 ANSWERS
• Homework Answers:
– Pg. 705, 15 - 22.
15. 3 ways
16. 15 ways
17. 40 ways
18. 1512 ways
19. 17,576,000;
11,232,000
20. 45,697,600;
32,292,000
21. 6,760,000; 3,276,000
22. 118,813,760;
78,936,000
55. 480
56. 2772
57. 2,176,782,336;
1,402,410,240
SCIENTIFIC INQUIRY AND ANALYSIS
75
HOMEWORK UNIT 4-2 ANSWERS
Pg. 705
23. 40,320
24. 120
25. 3,628,800
26. 362,880
31. 6
32. 20
33. 2
34. 5040
Pg. 706
60. 3,628,800
61.a. 720
b. 60,480
62. 210
63.12,612,600
Pg. 712
22. 495
23. 1
24. 3003
SCIENTIFIC INQUIRY AND ANALYSIS
25. 165
26. 792
27. 48
28. 778,320
Pg. 713
48. 1365
49. 315
52. 21,700
56. 210
76
HOMEWORK UNIT 4-3 ANSWERS
Pg. 719
21. 0.5
22. 0.9231
23. 0.23
Pg. 720-721
30. 0.0218
31. 0.455
32. 0.477
33. 0.545
34. 0.262
35. 0.0385
36. 0.154
39. 5.6 x 10-8
40. 0.001
43. 0.0527
SCIENTIFIC INQUIRY AND ANALYSIS
77
HOMEWORK UNIT 4-4 ANSWERS
Pg. 719-721
24. Exp = 21.7%
Th = 16.7%
25. Exp = 30.8%
Th = 33.3%
26. Exp = 49.2%
Th = 50%
27. Exp = 50.8%
Th = 50%
37. 26.2%
38. 1.285 x 10-10
41. a. P(tv) = 55.5%
b. P(vid) = 3.8%
42. a. P(loss) = 23.5%
b. P(gains) = 64.7%
46. 4.0%
SCIENTIFIC INQUIRY AND ANALYSIS
78
PROBABILITY
DO NOW: The Supreme Court of the US has 3
women and 6 men justices. On a certain case the
justices voted 5 to 4 in favor of the defendant.
1. What is the probability that exactly 2 out of 3
women voted in favor of the defendant?
2. What is the probability that at least 2 out of 3
women voted in favor of the defendant?
SCIENTIFIC INQUIRY AND ANALYSIS
79
HOMEWORK UNIT 4-5 ANSWERS
Pg. 727
4. 0.5
6. 0.5
8. 0.6
10. 0.1
12. 0.5
14. 0.67
16. 0.25
17. 0.7
18. 0.45 (yes)
25. 0.66
27. 0.25
Pg. 728
42. 0.83
43. 0.75
46. 0.1
47. 0.1
SCIENTIFIC INQUIRY AND ANALYSIS
80
PROBABILITY
DO NOW: Allergic reactions to poison ivy can be miserable. Plant oils cause the reaction.
A study was performed to see the effects of washing the oil off within 5 minutes of exposure.
A random sample of 1000 people with known allergies to poison ivy participated in the
study. Oil from the poison ivy plant was rubbed on a patch of skin. For 500 of the subjects,
the oil was washed off within 5 minutes. For the other 500 subjects, the oil was washed off
after 5 minutes. The results are in the table:
SCIENTIFIC INQUIRY AND ANALYSIS
81
HOMEWORK UNIT 4-6 ANSWERS
Pg. 734 &
735
12. 0.0468
13. 0.0468
18.a. 0.0625
b. 0.0637
19.a. 0.0059
b. 0.0060
25. 0.00144
26. 0.606
33. 0.581
SCIENTIFIC INQUIRY AND ANALYSIS
82
PROBABILITY
• DO NOW Problem
– Have out your homework.
– Do the following problem:
Radio station call letters (i.e. WNBC) consists of four
letters beginning with either K or W. How many different
radio station call letters are possible if letters can be
repeated? How many different radio station call letters are
possible if letters cannot be repeated?
SCIENTIFIC INQUIRY AND ANALYSIS
83
PROBABILITY
Evaluate the following:
a)
b)
c)
d)
e)
f)
4! =
12! =
3P 3 =
5P 1 =
10P1 + 5P2 =
4P 3 x 3P 2 =
1.
You
are
taking
a
Chemistry test and are
given a list of 10
elements. You are to
arrange the ten elements
in order as they appear
in the periodic table of
elements. Suppose you
have no idea of the
correct order and simply
guess.
How many
different ways can the
ten elements be listed?
2.
You
are
taking
a
Chemistry test and are
given a list of 10
elements. You are told
three of those elements
are elements 5, 6 & 7 on
the periodic table of
elements. You are asked
to pick elements 5, 6 & 7
and to arrange the three
elements in order as
they appear in the
periodic
table
of
elements. Suppose you
have no idea of the
correct elements or the
correct order and simply
guess.
How many
different ways can the
three elements be listed?
Suppose you know the
three elements but you
have no idea of the
correct order and simply
guess.
How many
SCIENTIFIC INQUIRY AND ANALYSIS
different ways can the
three elements be listed?
3.
A student council is
made up of 2 Freshmen,
2 Sophomores, 2 Juniors
and 3 Seniors. There are
75
Freshman,
68
Sophomores, 71 Juniors
and
82
Seniors.
Selections for each class
are
done
randomly
where the first pick of
the class is the active
council member and
subsequent picks are the
alternates. The senior
class has two alternates
and all of the other
classes have only one
alternate.
How many
different
student
councils can be formed?
84
PROBABILITY
1.
Evaluate the following:
a)
b)
c)
d)
e)
f)
2.
3.
class were chosen, how many
ways can they be selected?
3C3
=
C
1000 25 =
4C3 x 6C2 =
5C3 x 2C2 =
5C1 + 8C2 =
5P1 + 8P2 =
Junior Students
12
36
•
8
Seven cards are dealt from a
regular 52 card deck. How
many different ways can the
seven cards be dealt?
14
To the right is an indication of
how many students from the
junior class are taking AP
classes.
Three students are4.
picked from the junior class
(order doesn’t matter). How
many ways can those 3 students
be chosen from the whole junior
class? If only those students
who took just 1 AP class were
chosen, how many ways can
they be selected? If only those
students who took at least 1 AP
Took 1 AP Class
Took 2 AP Classes
Took 3 AP Classes
Took no AP Classes
& 2nd places), but it does not
matter in what order they finish.
How many quinella bets are
possible in this race?
EXTRA: If you make 4 exacta
bets in the race, what is the
probability you select the correct
one? If you make 4 quinella bets
in the race, what is the probability
you select the correct one?
There are 10 horses in a race.
One of the bets in a horse race is
the “exacta” in which you select
the 2 horses for “win and place”
(1st & 2nd places) in that exact
order. How many exacta bets
are possible in this race? The
“quinella” bet is similar to the
“exacta” in which you select the
2 horses for “win and place” (1st
SCIENTIFIC INQUIRY AND ANALYSIS
85