WEEK 2 Flashcards
(30 cards)
SIMPLE Probability (p)
(P/wanted) = n wanted divided by N of all possible
Conditional Probability
P (A GIVEN B) IS THE POSSIBILITY A to happen, Given that B has already happened
P( A AND B) = P (A GIVEN B) X P(B)
P(B GIVEN A ) = A AND B) / P (A)
Bayes theory
P (A GIVEN B) X P(B) = P(B GIVEN A) X P(A)
IF A, B IS INDEPENDENT THEN
P( A and B) or P (B and A) = P (A) X P (B)
IF A, B IS MUTUALLY EXCLUSIVE
P (A and B) = 0
then P(A OR B) = P(A) + P(B)
EXPECTATION = EPX
expected sum of each scenario (outcome) x each own probability outcome
if 2 scenarios given, then the two of them minus 1 is the 3rd
P(AnB) = P(A GIVEN B) X P(B) BUT
P(B AND A) = P (B GIVEN A) X P (A)
Permutations and Combinations
if order of selection is important, we use this; if it isn’t, then we use combinations
R = ITEMRS groups of size = n
With replacement/repetition, eg pin code
P(n,r) = n and 2 squared
withouth repetition, eg bookshelf
n Pr = n! / (n-r)!
Combinations
with repetition, eg ice cream shop
C(n,r) = (r+n-1)! / r!(n-1)!
without repetition, eg lottery
nCr = n!/ r!(n-r)!
Example: probability of winning a lottery?
to win, you need 6 numbers guessed correctly from 1 to 49
n = 49
r=6
nCr = 49! / 6!x(49-6)!
CERTAIN = 1
IMPOSSIBLE = 0
5 SCALES OF 0 TO 1
IMPOSSIBLE
UNLIKELY
EQUALLY LIKELY
LIKELY
CERTAIN
Classical apprach
events assigned (sometimes equally) probabilities based on theoretical outcomes
Empirical or relative frequency approach
probabilities retroactively assigned throught experiments = nobody knows the probabilities, but we experiment again and again to obtain outcomes and look at the frequency of them happening
Subjective approahc
qualification of personal belief assigned based on it
classical approach = What is the probability of obtaining an odd number score on one throw of a die?
3/6 = 0.5
empirical approach = P(A) requires past or experimental data
number of times even as A occurred / total number of observations
example of emperical approach Out of 200 light bulbs manufactured on a particular day, 15 were faulty
P(light bulb manufacture on given day is faulty) = 15/200 = 0.075
subjective approach = a person uses judgment to assess the probability
i estimate that there is 0.9 probability that this product will make a profit next year
probability of 0.5 = 50 %
probability = 0 to 1
% = 0% to 100%
Indipended even example
P (it will rain in Bristol on a certain day)
= is independent of whether that day is a weekday or not, but is not independent of what month that day occurs, as some months are rainier than others
Conditonal probability example
a supermarket manager might judge
P(delivered goods are in a good condition) = 0.9
but
P(delivered goods are in a good condition given that packaging is damaged) = 0.3
MULTIPLICATION RULE CONT
If 2 events are dependent, then
P(A AND B) = P(A) X P(B GIVEN A)
