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summary of probability rules
basic terms to know:
outcome
event = a set of outcomes
sample space = the set of all possible outcomes
for k equally likely outcomes,
assign probability 1/k to each
for an event A consisting of m of these outcomes, assign probability P(A) = m/k
conditional probability P( B | A )
restricts the sample space to be the set A, a subset of the original sample space
thus, we are no longer considering all the outcomes in event B but only those outcomes in A that are also in B, or A ∩ B
for equally likely outcomes,
assign P(B | A) = number of outcomes in A ∩ B / number of outcomes in A
0 ≤ P(A) ≤ 1
P(A) = 1 if event A is logically certain
P(A) = 0 if event A is logically impossible
events related to A and B:
not A Ac^ , the complement of the set A
A or B equivalent to A ∪ B, the union of sets A and B
unless specifically stated, this is A or B or both
A and B both A and B
equivalent to A ∩ B, the intersection of sets A and B
saying A but B, for contrast, is the same as A and B
P( A ∪ B ) = P(A) + P(B) – P(A ∩ B) called the Addition Rule
important special case of the addition rule:
the sample space = A ∪ A c and these are non-overlapping,
so 1 = P(A ∪ Ac^ ) = P(A) + P(Ac^ )
from which we can show that P(Ac^ ) = 1 – P(A) probability of A complement
P( B | A) = P( A ∩ B ) / P( A )
if P( A ) = 0, then we define P( B | A ) = 0
P(A ∩ B) = P(A) P(B|A) called the Multiplication Rule
also = P(B) P(A|B)
from which we derive
P(B) P(A|B) = P(A) P(B|A)
and so P(A|B) = P(A) P(B|A) / P(B)
and P(B|A) = P(B) P(A|B) / P(A)
these formulas, or alternate versions of them, are called Bayes' Rule
if P( B | A ) and P( B | A c ) turn out to be equal to P(B),
events A and B are said to be independent events
and the Multiplication Rule simplifies to P(A ∩ B) = P(A) P(B)
P(A) = P(A ∩ B) + P(A ∩ B
c ) called Law of Total Probability
more generally, if the sample space = B 1 ∪ B 2 ∪ ... ∪ B (^) j with the B (^) i non-overlapping
then P(A) = ∑ =
j i 1 P( A^ ∩^ B^ i^ )
for any sets A and C, C = (A ∩ C) ∪ (A c ∩ C )
so we can show P(A | C) + P(A c | C) = 1
and P(Ac^ | C) = 1 – P(A | C)
there is no rule relating P( B| A c^ ) and P( B | A )
