1. Counting Principles (Permutations and Combinations )

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3. Example
A combination lock can be set to open to any 4-digit sequence. (a) How many sequences are possible? (b) How many sequences are possible if no digit is repeated?
Solution: (a) Since there are 10 digits namely 0, 1, 2…..9, there
are 10 choices for each of the digit. By the multiplication
principle, there are 10 ∙10 ∙10 ∙10 =10,000 different sequences.
(b)There are 10 choices for the first digit. It cannot be used
again, so there are 9 choices for the second digit, 8 choices for
the third digit, and then 7 choices for the fourth digit.
Consequently, the number of such sequences is
10 ∙ 9 ∙ 8 ∙7 =5040 different sequences.
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4. Example
How many different ways can you choose a bagel, muffin or donut to eat and coffee or juice to drink?
Tree diagram
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5. Example
A teacher is lining up 8 students for a spelling bee. How many different line-ups are possible?
Solution: Eight choices will be made, one for each space that will
hold a student. Any of the students could be chosen for the first
space. There are 7 choices for the second space, since 1 student
has already been placed in the first space; there are 6 choices for
the third space, and so on.
By the multiplication principle, the number of different possible
arrangements is 8 ∙7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙1 = 40,320.
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6. (order does matter) TI-83/84 function
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7. Example
A teacher wishes to place 5 out of 8 different books on her shelf. How many arrangements of 5 books are possible?
Solution : The teacher has 8 ways to fill the first space, 7 ways
to fill the second space, 6 ways to fill the third, and so on…
Since the teacher wants to use only 5 books, only 5 spaces can
be filled (5 events) instead of 8, for 8 ∙7 ∙ 6 ∙ 5 ∙ 4 = 6720
arrangements.
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8. Example
Find the number of permutations of the letters L, M, N, O, P, and Q, if just three of the letters are to be used. Solution:
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9. How many permutations are there of two letters from the set {A,B,C}.
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10. (order does not matter)
TI-83/84 function
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11. Example
How many committees of 4 people can be formed from a group of 10 people? Solution: A committee is an unordered group, so use the combinations formula for C(10,4).
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13. Example
From a class of 15 students, a group of 3 or 4 students will be selected to work on a special project. In how many ways can a group of 3 or 4 students be selected? Solution: The number of ways to select group of 3 students from a class of 15 students is C(15, 3) = 455. The number of ways to select group of 4 students from a class of 15 students is C(15, 4) = The total number of ways to select a group of 3 or 4 students will be the 1820.
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15. Example
(a) How many 4-digit code numbers are possible if no digits are repeated? Solution: Since changing the order of the 4 digits results in a different code, permutations should be used. (b) A sample of 3 light bulbs is randomly selected from a batch of 15. How many different samples are possible? Solution: The order in which the 3 light bulbs are selected is not important. The sample is unchanged if the items are rearranged, so combinations should be used.
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16. Example
(c) In a baseball conference with 8 teams, how many games must be played so that each team plays every other team exactly once? Solution: Selection of 2 teams for a game is an unordered subset of 2 from the set of 8 teams. Use combinations again. (d) In how many ways can 4 patients be assigned to 6 different hospital rooms so that each patient has a private room? Solution: The room assignments are an ordered selection of 4 rooms from the 6 rooms. Exchanging the rooms of any 2 patients within a selection of 4 rooms gives a different assignment, so permutations should be used.
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17. Introduction to Probability

18. Set – a collection of elements that satisfy a certain condition.
Write the elements belonging to the set {x | x is a state whose name begins with the letter O}.
Solution: The state names that begin with the letter O make up
the set {Ohio, Oklahoma, Oregon }.
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19. Decide if the statement is true or false.
Example 1
Decide if the statement is true or false.
Solution: The first set is a subset of the second because each element of first set belongs to second set. (The fact that the elements are listed in a different order does not matter.) Therefore, the statement is true.
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21. A) What is the sample space?
Sample Space – The set of all possible outcomes of an experiment or event.
A) What is the sample space?
B) What is the probability of getting a 3?
C) What is the probability of getting an odd number?
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22. Example
Two coins are tossed, and a head or a tail is recorded for each coin. Write the event E: the coins show exactly one head. Solution: Tossing a coin is made up of the outcomes heads (H) or tails (T). If S represents the sample space of tossing two coins, then S = {HH, HT, TH, TT}. Two outcomes satisfy this condition: so, E = {HT, TH}.
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23. Example Number of elements in the sets
Two coins are tossed, and a head or a tail is recorded for each
coin.
Give a sample space for this experiment.
What is the probability of getting at least one head.
Solution: A) Tossing a coin is made up of the outcomes heads (H)
or tails (T). If S represents the sample space of tossing two
coins then S = {HH, HT, TH, TT}.
Solution : B) E = {HH,HT,TH}
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25. Monty Hall probability puzzle

26. Probability of Multiple Events

27. (E’ is called the complement of E)
(no outcomes in common; if one event occurs the other cannot) 4 1 6 5 2
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28. Example
A study of workers earning the minimum wage grouped such workers into various categories, which can be interpreted as events when a worker is selected at random. Source: Economic Policy Institute. Consider the following events: E: worker is under 20; F: worker is white; G: worker is female. Describe the following event in words: Solution:
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29. Read as; the probability of event F occurring given E has occurred
Independent Events
The fact that one event has occurred, does not change the probability of the second event occurring.
** Events that are mutually exclusive must be dependent, but dependent events are not necessarily mutually exclusive**
Read as; the probability of event F occurring given E has occurred
Read as; the probability of event E occurring given F has occurred
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30. (General rule of addition)
For mutually exclusive events,
So the equation simplifies to
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31. Example
If a single card is drawn from an ordinary deck of cards, find the probability of an ace or a club. Solution: Let A represent the event “an ace ” and C the event “club card.” There are 4 aces in the deck, so There are 13 clubs in the deck, so Since there is 1 ace of club in the deck, By the union rule, the probability of the card being an ace or a Club card is
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32. Example
Suppose two fair dice are rolled. Find the probability that the sum is 8, or both die show the same number. Solution:
1-1
1-2
1-3
1-4
1-5
1-6
2-1
2-2
2-3
2-4
2-5
2-6
3-1
3-2
3-3
3-4
3-5
3-6
4-1
4-2
4-3
4-4
4-5
4-6
5-1
5-2
5-3
5-4
5-5
5-6
6-1
6-2
6-3
6-4
6-5
6-6
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33. Example
Find the probability that when two fair dice are rolled, the sum is less than 11. Solution: To calculate this probability directly, we must find the probabilities that the sum is 2, 3, 4, 5, 6, 7, 8, 9 or 10 and then add them. It is much simpler to first find the probability of the complement, the event that the sum is greater than or equal to 11.
1-1
1-2
1-3
1-4
1-5
1-6
2-1
2-2
2-3
2-4
2-5
2-6
3-1
3-2
3-3
3-4
3-5
3-6
4-1
4-2
4-3
4-4
4-5
4-6
5-1
5-2
5-3
5-4
5-5
5-6
6-1
6-2
6-3
6-4
6-5
6-6
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34. (General rule of multiplication)
When events are Independent;
So the equation simplifies to:

35. 1/6 9/34 0.54 2/7 9/25 Independent Not Independent Not Independent
Not mutually exclusive
mutually exclusive
Not mutually exclusive

36. P(F); represented by large circle.
P(E and F); represented by overlapping section
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37. Example
The following table shows national employment statistics. Use the table to find each probability.
9729
9392
8937
5801
4383
19121
7. P(male | professional)
8. P(laborer | female)
9. P(female | sales)
10. P(professional | female)
11. P(sales | male)
12. P(male | laborer)

38. C is the event careful driver. NA is the event no accident 0.80
Example
Use a tree diagram to solve the following problem.
16. A car insurance company compiled the following information from a recent survey. 75% of drivers carefully follow the speed limit Of the drivers who carefully follow the speed limit, 80% have never had an accident. Of the drivers who do not carefully follow the speed limit, 65% have never had an accident.
What is the probability that a driver does not carefully follow the speed limit and has never had an accident?
C is the event careful driver.
NA is the event no accident
0.80
0.75
0.20
0.65
0.25
0.35