Post on 20-Jan-2016
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 thenp(A) = the sum of the probabilities of all
elements of A0 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} 1/2
• Let B = {5,6} +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
• p(A Ac) = p(A) + p(Ac)
• 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)
{1,2,3,4,5,6}2
(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
• p(Ac Ac) = p(Ac) p(Ac) = (5/6) (5/6) = 25/36• (Ac Ac)c is the event of rolling two dice and
getting a 6 on at least one roll• p((Ac Ac)c) = 1 – 25/36 = 11/36
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
• AcAcAc = (Ac)3 is the event of rolling three dice and not getting a 6 on any of the rolls
• p((Ac)3) = (p(Ac))3 = (5/6)3 = 125/216• (AcAcAc)c is the event of rolling three dice
and getting a 6 on at least one roll• p((AcAcAc)c ) = 1 – 125/216 = 91/216 0.421
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