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Chapter 3
3.1
Sample Space = the set of all possible outcomes {1,2,3,4,5,6}
Event = a subset of the sample space (all evens)
Outcome = the result of a single trial
Probability = the chance that something will occur.
P(E) = probability of event E
probability is a number from 0 to 1. 0 ≤ P(E) ≤ 1
P(E) = 0 means the event can never occur
P(E) = 1 means the event will occur for sure
Three types of probability:
Classical probability = theoretical probability =
number of desired outcomes
total of all possible outcomes
calculate from known possible outcomes.
Empirical probability = experimental probability =
actual number of desired outcomes
get data from observation or experiment or survey.
total of all trials
Subjective probability = intuition = guess. Not based on any math or data.
Tree Diagram (for multiple events in sequence)
Use for rolling a die or tossing a coin multiple times, for dinner options, for clothing options.
Example: Toss a coin 3 times:
H
HHH
T
H
HHT
HTH
T
HTT
H
THH
T
H
THT
TTH
T
TTT
H
H
T
start
H
T
T
8 possible outcomes
Sample Space {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Toss two dice: 36 possible outcomes
Sample space {11 12 13 14 15 16 21 22 23 24 25 26 31 32 33 34 35 36
41 42 43 44 45 46 51 52 53 54 55 56 61 62 63 64 65 66}
Deck of cards: 52 possible outcomes
Fundamental Counting Principle instead of counting them all, multiply their numbers
for independent events in sequence, multiply the choices of each.
Example 1: toss a coin 3 times. 2 x 2 x 2 = 8 possible outcomes
Example 2: toss 2 dice. 6 x 6 = 36 possible outcomes
Example 3: 2 appetizers, 3 main courses, 4 desserts.
2 x 3 x 4 = 24 different dinner menus
Example 4: 3 shirts, 2 pants, 2 pair of shoes. 3 x 2 x 2 = 12 different clothing outfits
Example 5: 3 digit lock combination code 10x10x10 = 1000 different lock combinations
Example 6: license plate of 3 letters followed by 3 numbers
26x26x26x10x10x10 = 17576000
Example 7: 7 digit telephone number, first digit not zero. 9x10x10x10x10x10x10=9000000
Example 8: Toss 1 coin and 1 die. 2x6 = 12 different outcomes
Example: 10 multiple choice questions, 4 choices each.
How many different ways can the test be answered? 4x4x4 ... x4 = 4^10 = 1048576
With repeat = with replacement. the digit or card can be repeated. Card returned to deck.
Without repeat = without replacement. after a choice is used once, it cannot be used again,
card not returned to deck.
Example: license plate of 6 letters with no repeats. 26x25x24x23x22x21 = 165765600
Law of Large Numbers: When an experiment is repeated many, many times,
the empirical probability approaches the theoretical probability of the event.
Example: if you toss a coin 10,000 times, P(heads) will be near to 0.5.
Frequency Distributions and Probability
Age
frequency
15 to 24
54
25 to 34
366
35 to 44
233
45 to 54
180
55 to 64
125
65 and over
42
total
1000
P(age 25-34) = 366/1000= 0.366
same as relative frequency
Complementary events = opposites, add to 100%
E = event
E ' = complementary event (not E)
P(E) + P(E ') = 1.00
P(E ') = 1 - P(E)
P(rain) + P(not rain) = 1.00
P(rain) = .35 P(not rain) = 1 - .35 = .65
E
E'
3.2
Multiplication Rule = multiply probabilities when two or more events in sequence
Example: roll a die twice. Probability of rolling two sixes = 1/6 x 1/6 = 1/36
1 1
1
Example: roll a die and toss a coin. Probability of getting a tail and rolling a 2 = 2 𝑥 6 = 12
be careful of independent and dependent events
Conditional Probability = probability of an event occurring, given that another event has already
occurred.
given that
P(second event | first event that already occurred)
Example: P(shoveling | it snowed)
Example: P(queen | diamond) = 1/13
Example: P(jack | face card) = 4/12 = 1/3
Example: P(2nd card is queen | first card is king) = 4/51
Male
Female
Total
Owns a cat
54
23
67
Does not own a cat
22
7
29
Total
76
30
96
Example:
Example:
Example:
P(owns cat | female) = 23/30
P(male | owns a cat) = 54/67
P (does not own a cat | male) = 22/76
Independent Event = is not influenced by another event, outcome is not related to another event
Example: coin, die, marbles in a hat, cards
Dependent Event = is influenced by another event
Example: rain today (dependent on weather yesterday as weather has weather patterns)
Example: people behavior is often not random chance, opinions, actions
Live in city
Live in suburb
total
Age 20 to 30
10
5
15
Age 30 to 40
20
20
40
Age 40 to 50
15
15
30
Over age 50
5
10
15
Total
50
50
100
If living in the city or suburb is independent of age, then the ratios will be the same for all ages.
Overall for all ages. Living in city = 50/100 = 1/2
Age 20 to 30. living in city = 10/15 = 2/3
Age 30 to 40. living in city = 20/40 = 1/2
Age 40 to 50. living in city = 15/30 = 1/2
Over 50.
living in city = 5/15 = 1/3
These are not the same ratios, so age and living in city is not independent. There is some relationship
between age and living in city, some dependency.
Without replacement = dependent events, next item selected is influenced by the first
With replacement = independent events, next item selected is not influenced by the first
If truly independent, then P(B) = P(B|A).
The fact that A event occurred has no influence on B occurring.
Multiplication rule with independent events
P(A and B) = P(A) * P(B)
Example: Probability of drawing 2 queens with replacement
independent, first card does not influence second card
P(Q and Q) = P(Q) * P(Q) = 4/52 * 4/52 = 16/2704 = 1/169 = .0059
Mulitiplication rule with dependent events
(A and B) = P(B) * P(A|B)
The probability of A and B occurring = probability of B * probability of A occurring when we know B
has already occurred.
Example: Probability of drawing 2 queens without replacement
Dependent (the second card probability is different with one card already missing)
P(Q and Q) = P(Q) * P(Q|Q) = 4/52 * 3/51 = 12/2652 = .0045
Prob of drawing 2 queens with replacement = probability of getting a queen (4/52) * probability of
getting a queen on the second card when we know the first card is a queen (3/51)
3.3
P(A and B) = P (both events occurring)
P(A or B) = P(either event or both occurring)
Addition Rule
P(A or B) = P(A) + P(B) – P(A and B)
probability of A or B (or both) = prob of A + prob of B – overlap, if any
Mutually Exclusive Events = events that cannot happen at the same time
For mutually exclusive events, P(A and B) = 0
When mutually exclusive, you can just add the probabilities
Example: Rolling a 3 and a 4 on the same roll of a die. P(3 and 4) = 0
Use adding rule for mutually exclusive events (no overlap):
P(3 or 4) = P(3) + P(4) – P(3 and 4)
= 1/6 + 1/6 –
0
= 2/6 = 1/3
Use adding rule for not mutually exclusive events (has overlap):
Example: Roll a die, find probability that it is odd or less than 4.
P(odd or less than 4) = P(odd) + P(less than 4) – P(odd and less than 4)
{1,2,3,5}
{1,3,5} {1,2,3}
{1,3}
= 3/6
+
3/6
2/6
= 4/6 = 2/3
History class
Chemistry class
Yes
No
Total
Yes
27
33
60
No
15
25
40
Total
42
58
100
P(history or chemistry) = P(history) + P(chemistry) – P(history and chemistry)
= 42/100 + 60/100 - 27/100
= .42
+ .60
- .27 = .75
Remember P(A or B) means prob of having either A or B or BOTH occur.
Venn Diagram
history
chemistry
33
15
27
25
3.4
Permutations = total number of ways or arrangements when order matters (larger number)
Combinations = number of different arrangements when order doesn't matter (smaller number)
Putting n items in order or in a line = n!
When order matters
Choosing r items from a group of n = nPr = n choose r =
𝑛!
(𝑛−𝑟)!
examples: choosing first, second, third place
choosing pres, vice pres, secretary, treasurer
choosing a 3 digit code (each digit is 0 to 9)
When order doesn't matter
Choosing r items from a group of n items = nCr = n choose r =
𝑛!
𝑟!(𝑛−𝑟)!
Examples: choosing a committee
choosing representatives to send to a meeting
choosing 5 CD's to take to a party from your collection of 50
hiring 4 new employees out of 20 applicants – all positions are equal
choosing 2 people to interview from 12 jurors
choosing 3 pizza toppings out of 9 possibilities
choosing 2 ice cream flavors out of 33 possibilities
Distinguishable ways
when some items repeat we need to divide by the items that are “indistinguishable”
Example: plant 20 trees on a street in a line evenly spaced. 6 oak, 9 maple, 5 poplar.
In how many distinguishable ways can the trees be planted?
20!
6!9!5!
= 77597520
Probability using number of possibilities
number of ways the desired outcome A can happen
P(A) =
total number of ways
Example: Find the probability of getting 5 diamonds in a 5 card hand.
=(Ways of choosing 5 diamonds from 13 diamonds) / (total number of 5 card hands from 52 cards)
13 C 5
1287
= 52 𝐶 5 = 2598960 = 0.000495