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