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Unformatted text preview: omes •Consider a more complicated example: rolling 2 dice. What is P({2,6})? It is the same as P({1,1}) or
P({3,4}). All outcomes are equally likely. We need to
count the outcomes or have rules to compute
probabilities. Events ORF 245 Prof. Rigollet
Fall 2012 An event is a collection of outcomes. It can be
described either with words or using formal notation
from set theory. Passing from the ﬁrst one to the
second is a necessary skill.
Flipping two coins. We know that the outcomes are
(H,H), (H,T), (T, H), (T,T). Consider the events
{twice the same}={(H,H), (T,T)}
{heads ﬁrst}={(H,T), (H,H)}
{no heads}={(T,T)}
We want to ﬁnd rules to compute the probability of
events from the probability of outcomes. ORF 245 Prof. Rigollet
Fall 2012 Operations on events B A Operations on events ORF 245 Prof. Rigollet
Fall 2012 Union of two events A and B:
A B = {outcomes that are either in A or in B or in both} Intersection of two events A and B:
A B= {outcomes that in A and in B} Complement of the event A:
Ac = {outcomes that are not in A} ORF 245 Prof. Rigollet
Fall 2012 Operations on events Union B A B A “or” B
A[B A Intersection A B B A “and” B
A Complement Ac “not” A A\B A
A c Operations on events Flipping two coins.
A={twice the same}={(H,H), (T,T)}
B={heads ﬁrst}={(H,T), (H,H)}
C={no heads}={(T,T)}
A B ={(H,H), (T,T), (H,T)} ={twice the same or heads ﬁrst}
A B ={(H,H)} ={twice the same and heads ﬁrst}
={twice heads}
C c ={ }
={at least one heads} ORF 245 Prof. Rigollet
Fall 2012 Addition rule of
disjoint events ORF 245 Prof. Rigollet
Fall 2012 Two events are disjoint if they have no outcome in
common. For example:
{(H,H), (T,T)} and {(H,T), (H,T)} are disjoint
{(H,H), (T,T)} and {(H,T), (H,H)} are not disjoint
Equivalently, two events are disjoint if their
intersection is empty (no outcome is in in both)
For disjoint events A and B we have the addition
rule
P (A B ) = P (A) + P (B ) ORF 245 Prof. Rigollet
Fall 2012 Addition rule of
disjoint events More generally, if k events A1 , A2 , . . . , Ak are disjoint
P (A1 A2 ... Ak ) = P (A1 ) + P (A2 ) + . . . + P (Ak ) B
A A B C
D P (A B ) = P (A) + P (B ) P (A B C D) = P (A) + P (B ) + P (C ) + P (D) Addition rule of
disjoint events ORF 245 Prof. Rigollet
Fall 2012 Example: Tossing ﬁve coins at random. What is the
probability of at least four heads?
Possible outcomes={HHHHH, THHHH, HTHHH,
HHTHH, HHHTH, HHHHT, TTHHH,...,TTTTT}
There are 25 possible outcomes (2 for each round)
P(at least 4 heads) =P(4 heads or 5 heads)
=P(4 heads)+P(5 heads)
=1/32 + 5/32 = 6/32=3/16 Ge...
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This document was uploaded on 10/14/2013.
 Fall '09

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