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} 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 A c are disjoint, so p(A  A c ) = p(A) + p(A c ) p(S) = p(A) + p(A c ) 1 = p(A) + p(A c ) 1  p(A) = p(A c ) p(A c ) = 1  p(A) Also, p(A) = 1  p(A c )

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} A c = S  {6} = {1,2,3,4,5,6} – {6} = {1,2,3,4,5} p(A) = 1/6 P(A c ) = 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 A c is the event of rolling one die and not getting a 6. p(A c ) = 1 – p(A) = 5/6 A c  A c is the event of rolling two dice and not getting a 6 on either roll p(A c  A c ) = p(A c )  p(A c ) = (5/6)  (5/6) = 25/36

Probability of A  B Then A c is the event of rolling one die and not getting a 6. p(A c ) = 1 – p(A) = 5/6 A c  A c is the event of rolling two dice and not getting a 6 on either roll p(A c  A c ) = p(A c )  p(A c ) = (5/6)  (5/6) = 25/36 (A c  A c ) c is the event of rolling two dice and getting a 6 on at least one roll p((A c  A c ) c ) = 1 – 25/36 = 11/36

Probability of A  B Then A c is the event of rolling one die and not getting a 6. p(A c ) = 1 – p(A) = 5/6 A c  A c  A c = (A c ) 3 is the event of rolling three dice and not getting a 6 on any of the rolls p((A c ) 3 ) = (p(A c )) 3 = (5/6) 3 = 125/216 (A c  A c  A c ) c is the event of rolling three dice and getting a 6 on at least one roll p((A c  A c  A c ) 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