Chapt 1

Question 0

Probability

Answer 0

The extent to which something is likely to happen

Question 1

Experiment

Answer 1

any process that requires some action to be performed and has an outcome that can be recorded.

Question 2

Outcome

Answer 2

Any single result from an experiment

Question 3

Sample space

Answer 3

The set/collection of all possible outcomes of an exoeriment and is typically denoted as S
Eg s={ 1 2. 3}

Question 4

Event

Answer 4

Any collection of outcomes from the sample soacmore formally in terms of set theory it is a subset of the sample space
Usually denoted by capital letters

Question 5

The null event

Answer 5

Or empty set contains no outcomes and is denoted by ∅.

Question 6

Intersection A ∩ B

Answer 6

Means both A and B occur “∩” denotes logical AND

Question 7

Union A∪B

Answer 7

Either A or B or both occur U denotes logical OR

Question 8

Complement A ̄ ( bar on top of A)

Answer 8

Not A which occurs when A does not,

Sometimes denoted Ac and is the logical NOT

Question 9

Disjoint A∩B=∅

Answer 9

A and B have no points in common so if A occurs then B does not
AKA mutually exclusive

Question 10

Subset A ⊂ B

Answer 10

If A happens them B will definitely happen AKA A implies B

Question 11

Compound experiment

Answer 11

Large experiment made u if two or more experiments

Question 12

2^k

Answer 12

Total number of outcomes when tossing k coins

Question 13

6^k

Answer 13

Number of outcomes when tolling 6 dices

Question 14

Independent

Answer 14

Independence is when knowing the result of one event does not effect the probability of the other occurring my eg

P(A n B)= p(A) p(B)

Question 15

If A and B are independent…

Answer 15

Then P(A n B is equal to all of the following

P(A|B) P(B)
P(B|A) P(A)
P(A) P(B)

Question 16

three main methods of counting events

Answer 16

1 multiplication principle
2 permutations
3 conbinations

Question 17

the multiplication principle

Answer 17

Suppose there are two experiments 1, 2
which contain n1, n2 outcomes
Then the joint experiment of conducting 1 and 2 together has n1 x n2 possible outcomes.
if there were 3 experiments it would be n1 x n2 x n3 and so on…

Question 18

permutations

(orderings or objets

Answer 18

when there are n number of different objects and n spaces to put them in
so the number of ways of putting the n different objects in the n different spaces is…
n! = n x (n - 1) x (n- 2) x . . . x 2 x

Question 19

n! and 0!

Answer 19

0!=1
n!…
The 1st object can be put in any one of n spaces.
• The 2nd object can be put in any one of (n 1) spaces (one space is already taken).
• The nth object can only be put in 1 space.

Question 20

when the order in which you select the n objects is important

Answer 20

eg selecting AB is different fromBA...
then the number of possible permutations from n is:
nPr=  n!
        (n - r)!
***divided by***

Question 21

when the objects are not all different…

Answer 21

there are t < n di↵erent types. Assume there are n1 objects of type 1, n2 of type 2,. . . , nt objects of type t, where n1 + n2 + . . . + nt = n. Then the number of permutations (orderings) of all n objects is:
n!
n1! x n2! x… x nt!

** divided by**

Note that in the above formula if all objects are di↵erent then nr = 1 for all groups r and we obtain the same formula as above.

Question 22

combinations

Answer 22

Suppose there are a collection of n objects all of which are di↵erent. From these n, 0  r  n are chosen. Then if the order in which they are chosen does not matter, then there are
nCr = n!
(n r)!r!
…possibble combinations

it is written as n over r in a bracket without a line between them and is said as “n chooses r”

Question 23

probability rule 4:

the general addition law

Answer 23

P(A U B) = P(A) + P(B) P(A n B)

can also be…

P(A n B)=P(A)+P(B) P(A u B).

• P(A u B) = P(A) + P(B) if A and B are disjoint, i.e. if A n B = 0/

Question 24

probability rules 1: P(S) = 1

Answer 24

where S is the sample space (something must happen).

Question 25

probability rules 2: P(0/)=0 ** 0 with line through...empty set

Answer 25

where 0/ is the empty event

Question 26

probability rules 3: P(A ̄) = 1 P(A) **A ̄ MEANS AWITH BAR ONTOP

Answer 26

A or A ̄ must happen

Question 27

relative frequency

Answer 27

an experiment is repeated several times and the proportion of times that event A occurs is that the probability of it occurring in the future rfn= number of times A occurred/ number of repetitions of experiment= r/n

Question 28

subjective probability

Answer 28

when an individual estimates the probability of event A occurring based on theyr prior knowledge

Question 29

p (AlB)=

Answer 29

probability of A given that B has already happened

Question 30

conditional probability

Answer 30

the conditional probability that event A occurs given that B has already happened is P(AlB)= P (A n B)/ P(B) can also be written as P(AnB)= P(AlB)P(B)

Question 31

equally likely outcomes probability model

Answer 31

if an experimenter has an experiment with n outcomes that are all equally likely to occur the probability of any single outcome is P(any single outcome)=1/n then for an event A made up of m outcomes its probability is: P(A)=m/n= number of favourable outcomes /total number of outcomes

Question 32

general addition law

Answer 32

P(AUB)=P(A) + P(B)-P(AnB)

Question 33

independence

Answer 33

``` events A and B are independent if: P(AnB)=P(A)P(B) informally, they are independent if knowing the result of one event does not affect the probability of the other occurring if independant P(AnB) is equal to: P(AlB)p(B), PBlA)P(A), P(A)P(B) ```

Question 34

P(AlB) if A and B are disjoint?

Answer 34

PAnB)= PAlB)P(B) P(AlB)=P(AnB)/P(B) P(AlB=0/P(B) =0

Question 35

miltiplication principle

Answer 35

allows us to determine the number of outcomes in a compound experiment- that is one with several experiments multiply together number of outcomes e.g. rolling one dice and tossing a coin =12 outcomes in general number of outcomes to the power of how many times experiment is done so for coin it would be 2(outcomes) to the power of k (how many coins were tossed

Question 36

permutations

Answer 36

n different objects and n spaces to put them number of different ways to put them is n! this is called the number of permutations i.e. the number of different orders in which you can select the n objects

Question 37

permutations

Answer 37

when the order in which you select the n objects is important nPr= n!/(n-r)!

Question 38

say you have different numbers of the n objects then

Answer 38

the number of orderings is n!/n(type1)!+n(type2)!+n(type3)! | r= groups so number of different types

Question 39

permutation exampleyou have one suspect and eight normal people how many ways can you order them

Answer 39

8!=

Question 40

combinations

Answer 40

collection of n objects all of which are different from these r are chosen then if order they are chosen doesn't matter then nCr=n!/(n-r!)r! this is also written (n over r without a line in between) and its pronounced n choose r

Question 41

Bayes theory

Answer 41

Bayes theorem is based on the idea of a partition, which is a set of events that follow the three rules defined below. last ting in chapt 1 go through with examples******