Download Sample space (S) - Temple University

Probability
Prof. Richard Beigel
Math C067
September 27, 2006
Experiments
• An experiment is a process that does may
not always give the same result.
• Performing an experiment once is called a
trial.
• The result of a trial is called its outcome.
Probability spaces
• Sample point = outcome
• Event = a set of outcomes
• Sample space (S) = the set of all possible
outcomes (S is analogous to the universal
set U from the set-theory lectures)
• Disjoint events are called mutually exclusive
Probabilities
• If x is a sample point (outcome),
The probability of x is called p(x)
0  p(x)  1
• If A is an event then
p(A) = the sum of the probabilities of all
elements of A
0  p(A)  1
• p({}) = 0
• p(S) = 1
Single Fair Coin Flip
• S = {H,T}
• p(H) = ½
• p(T) = ½
Single Fair 6-Sided Die Roll
•
•
•
•
•
•
•
S = {1,2,3,4,5,6}
p(1) = 1/6
p(2) = 1/6
p(3) = 1/6
p(4) = 1/6
p(5) = 1/6
p(6) = 1/6
Soccer game
•
•
•
•
S = {Win,Lose,Tie}
p(Win) = ?
p(Lose) = ?
p(Tie) = ?
Equiprobable Outcomes
If all outcomes are equally likely (as with a
fair die or a fair coin) then
• p(x) = 1/|S|
• p(A) = |A|/|S|
Outcomes are not always equally likely, so
use these formulas with caution.
Single Fair 6-Sided Die Roll
A single fair 6-sided die is rolled.
• Let A be the event that an odd number is
rolled.
• A = {x  S : x is odd} = {1,3,5}
• p(A) = |A|/|S| = 3/6 = ½
Single Fair 6-Sided Die Roll
A single fair 6-sided die is rolled.
• Let B be the event that a number greater
than 4 is rolled.
• B = {x  S : x > 4} = {5,6}
• p(B) = |B|/|S| = 2/6 = 1/3
Single Fair 6-Sided Die Roll
A single fair 6-sided die is rolled.
• A  B is the event that an odd number
greater than 4 is rolled.
• A  B = {x  S : x is odd and x > 4} = {5}
• p(A  B) = |A  B|/|S| = 1/6
Single Fair 6-Sided Die Roll
A single fair 6-sided die is rolled.
• A  B is the event that a number that is odd
or greater than 4 is rolled.
• A  B = {x  S : x is odd or x > 4} =
{1,3,5,6}
• p(A  B) = |A  B|/|S| = 4/6 = 2/3
Probability of Union
• p(A  B) =? p(A) + p(B)
• Let A = {1,3,5}
• Let B = {5,6}
1/2
+1/3
• A  B = {1,3,5,6}
2/3
Probability of Union
•
•
•
•
•
p(A  B) = p(A) + p(B)  p(A  B)
Let A = {1,3,5}
1/2
Let B = {5,6}
+1/3
A  B = {5}
1/6
A  B = {1,3,5,6}
=2/3
Mutually Exclusive Events
• If A and B are mutually exclusive events, i.e.,
disjoint sets then
p(A  B) = p(A) + p(B)
• Why?
• Because A  B = {},
• p(A  B) = p(A) + p(B)  p(A  B)
•
= p(A) + p(B)  p({})
•
= p(A) + p(B)  0
•
= p(A) + p(B)
Complement
•
•
•
•
•
•
•
A and Ac are disjoint, so
c
c
p(A  A ) = p(A) + p(A )
p(S) = p(A) + p(Ac)
1 = p(A) + p(Ac)
1  p(A) = p(Ac)
p(Ac) = 1  p(A)
Also, p(A) = 1  p(Ac)
Single Fair 6-Sided Die Roll
A single fair 6-sided die is rolled.
• Let A be the event that a 6 is rolled
• A = {6}
• Ac = S  {6} = {1,2,3,4,5,6} – {6} = {1,2,3,4,5}
• p(A) = 1/6
• P(Ac) = 1 – 1/6 = 5/6
Rolling Two Dice
• Sample space = the set of all ordered pairs
of die rolls
• = {(x,y) : 1  x  6 and 1  y  6}
• = {1,2,3,4,5,6}  {1,2,3,4,5,6}
• = {1,2,3,4,5,6}2
• To save some writing we will write xy
instead of (x,y)
2
{1,2,3,4,5,6}
(Cartesian) Product of Two Sets
•
•
•
•
A  B = {(a,b) : a  A and b  B}
Let A = {egg roll, soup}
Let B = {lo mein, chow mein, egg fu yung}
AB=
{(egg roll,lo mein), (egg roll, chow mein),
(egg roll,egg fu yung), (soup,lo mein),
(soup,chow mein), (soup,egg fu yung)}
Rolling Two Dice
Probability of A  B
• Outcomes must be equiprobable
• P(A  B) = p(A)  p(B)
• Let A = the event of rolling one die and getting a 6.
p(A) = 1/6
• Then Ac is the event of rolling one die and not getting a
6. p(Ac) = 1 – p(A) = 5/6
• Ac  Ac is the event of rolling two dice and not getting a
6 on either roll
• p(Ac  Ac) = p(Ac)  p(Ac) = (5/6)  (5/6) = 25/36
Probability of A  B
• Then Ac is the event of rolling one die and not
getting a 6. p(Ac) = 1 – p(A) = 5/6
• Ac  Ac is the event of rolling two dice and not
getting a 6 on either roll
c
c
c
c
• p(A  A ) = p(A )  p(A ) = (5/6)  (5/6) = 25/36
c
c c
• (A  A ) is the event of rolling two dice and
getting a 6 on at least one roll
c
c c
• p((A  A ) ) = 1 – 25/36 = 11/36
Probability of A  B
• Then A is the event of rolling one die and not
getting a 6. p(Ac) = 1 – p(A) = 5/6
c
c
c
c 3
• A A A = (A ) is the event of rolling three
dice and not getting a 6 on any of the rolls
c 3
c 3
• p((A ) ) = (p(A )) = (5/6)3 = 125/216
c
c
c c
• (A A A ) is the event of rolling three dice and
getting a 6 on at least one roll
c
c
c c
• p((A A A ) ) = 1 – 125/216 = 91/216  0.421
c
Probability of A  B
• Suppose that we roll two dice.
• What is the probability that we get two 6s?
• Let A be the event of getting a 6 when we roll
one die
• P(A  A) = p(A)  p(A) = (1/6)(1/6) = 1/36
4 the hard way
• Suppose that we roll two dice.
• What is the probability that we get two 2s?
• Let A be the event of getting a 2 when we roll
one die
• P(A  A) = p(A)  p(A) = (1/6)(1/6) = 1/36
Probability of A  B
• Suppose that we roll two dice.
• What is the probability that we get a 1 on the first
die and a 3 on the second die?
• Let A be the event of getting a 1 when we roll one
die
• Let B be the event of getting a 3 when we roll one
die
• P(A  B) = p(A)  p(B) = (1/6)(1/6) = 1/36
• In fact each particular outcome has probability 1/36
4 the easy way
• Suppose that we roll two dice.
• What is the probability that one of the rolls is a 1
and the other is a 3?
• The event in question consists of two outcomes.
Let A = {(1,3),(3,1)}
• The sample space S = {1,2,3,4,5,6}2
• p(A) = |A|/|S| = 2/36 = 1/18
Probability of A  B
• Suppose that we roll two dice.
• What is the probability that the sum of the rolls is
7?
• Let A = {(x,y) : x+y = 7} =
{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}
• The sample space S = {1,2,3,4,5,6}2
• p(A) = |A|/|S| = 6/36 = 1/6
Probability of A  B
• Suppose that we roll two dice.
• What is the probability that the sum of the rolls is
4?
• Let A = {(x,y) : x+y = 4} = {(1,3),(2,2),(3,1)}
• The sample space S = {1,2,3,4,5,6}2
• p(A) = |A|/|S| = 3/36 = 1/12