unit 4 probability scientific inquiry and analysis1

SCIENTIFIC INQUIRY AND ANALYSIS 1

UNIT 4

PROBABILITY


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)


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)


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.


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.


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.


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?


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


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?


PROBABILITY

• Counting Example 1:

WHOLE WHEAT BREAD

HONEY HAM

SALAMI

BOILED HAM

PASTRAMI

ROAST BEEF

TURKEY

CHICKEN

BOLAGNA

SWISS CHEESE

PROVOLONE

AMERICAN CHEESE

CHEDDAR CHEESE

MUENSTER CHEESE

MAYO

OLIVE OIL

MUSTARD


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 . . .)


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


• If Jonah chooses one of each, how many different sandwiches can be made?


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.


WHOLE WHEAT BREAD

HONEY HAM

SALAMI

BOILED HAM

PASTRAMI

ROAST BEEF

TURKEY

CHICKEN

BOLAGNA

SWISS CHEESE

PROVOLONE

AMERICAN CHEESE

CHEDDAR CHEESE

MUENSTER CHEESE

MAYO

OLIVE OIL

MUSTARD

SALAMI

PASTRAMI

ROAST BEEF

TURKEY

CHICKEN

BOLAGNA

HONEY HAM

SALAMI

PASTRAMI

ROAST BEEF

TURKEY

CHICKEN

BOLAGNA

PROBABILITY

• Fundamental Counting Principle without Repetition


PROBABILITY

• Fundamental Counting Principle without Repetition– 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.


PROBABILITY

• Fundamental Counting Principle without Repetition– 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.


PROBABILITY

• Fundamental Counting Principle with Repetition– 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?


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?


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 . . .)


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?


PROBABILITY

• Factorials– Factorial (m!) is a mathematical operation that

multiplies a number m by this method :

– Example:

– Note: 1! = 1 & 0! = 1

𝑚× (𝑚−1 ) ×(𝑚−2)⋯×1


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.

• P(n,r) is sometimes phrased: “Permutations of n taken r times”


PROBABILITY

• Permutation properties


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?


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?


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?


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:

• However some of the outcomes are not indistinguishable.

MOM, OMM, MMO, MOM OMM, MMO• In this case only 3 outcomes are distinguishable.


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.


PROBABILITY

• Permutation with Repetition Example 1:– Find the number of distinguishable permutations

of the letters in REBECCA.


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.


PROBABILITY

• Combinations – how many different r objects out of n objects, no order.

• Given the same number of n objects and r objects, which is larger the combination or the permutation?


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.


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?


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?


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)


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. In how many different ways can you choose 4 pairs of shorts to take on vacation?

2. In how many different ways can you choose 2 shirts to wear on the 1st and 2nd days of vacation?

3. If you bring 4 pairs of shorts and 6 shirts, how many different outfits can you make?


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.

38

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).

SCIENTIFIC INQUIRY AND ANALYSIS

OUTCOMES

SAMPLE SPACE (S)

EVENT (E)


PROBABILITY

• Probability OUTCOMES

SAMPLE SPACE (S)

EVENT (A)

𝑃 ( 𝐴)= 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠𝑖𝑛 𝐴𝑡𝑜𝑡𝑎𝑙𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠


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) =


PROBABILITY

• Example using a Punnett Square


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) =


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) =

Thin Crust53%

Pan Pizza22%

Thick Crust23%

Stuffed Crust2%

Pizzas Sold


PROBABILITY

• Counting Principle Probability Example:– GP Clayton goes to a sandwich shop to order a


• If GP Clayton chooses one of each, how many different sandwiches can be made?

# of sandwiches = 5 x 8 x 5 x 3 = 600


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) =


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) =


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.


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.


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) =


PROBABILITY• Example of Telephone Numbers

EVENT NUMBER OF OUTCOMES

FREQUENCY = # OF OUTCOMES/SAMPLE SPACE

Telephone # ends in 1











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?


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


PROBABILITY

• Unions & Intersections Review


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)


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)


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)


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)


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) =


PROBABILITY

• Complements

– 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)

A’ A


PROBABILITY

• Complement Example 1:– In throwing two 6 sided dice what is the

probability that the sum is not 11?


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?


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?


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)


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.


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?


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?


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.


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)


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?


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?


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?


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)


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)


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?


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



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

25. 165

26. 792

27. 48

28. 778,320

Pg. 713

48. 1365

49. 315

52. 21,700

56. 210



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



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%


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


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

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:


Pg. 734 & 73512. 0.046813. 0.046818.a. 0.0625 b. 0.063719.a. 0.0059 b. 0.006025. 0.0014426. 0.60633. 0.581



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?


PROBABILITYEvaluate the following:

a) 4! =b) 12! =c) 3P3 =

d) 5P1 =

e) 10P1 + 5P2 =

f) 4P3 x 3P2 =

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 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?


PROBABILITY1. Evaluate the following:a) 3C3 =

b) 1000C25 =

c) 4C3 x 6C2 =

d) 5C3 x 2C2 =

e) 5C1 + 8C2 =

f) 5P1 + 8P2 =

2. Seven cards are dealt from a regular 52 card deck. How many different ways can the seven cards be dealt?

3. To the right is an indication of how many students from the junior class are taking AP classes. Three students are 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 class were chosen, how many ways can

they be selected?

4. 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 & 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?

36

14

8

12Junior Students

Took 1 AP Class Took 2 AP Classes

Took 3 AP Classes Took no AP Classes

unit 4 probability scientific inquiry and analysis1

Documents

different ways

pair of blue pants

probabilityscientific

number of possible outcomes

tree method

grid method

pair of black pants

red shirt