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Rules of Probability Axioms of Probability Recall: P [ E ] ≥ 0. P [ S ] = 1 if E i ∩ E j = f Property 3 is called the additive rule for probability for any event E Note: hence or Note: Example –Probabilities for various events in Lotto 6/49 Let E = Event that no money is won
E N D
Axioms of Probability Recall: • P[E] ≥ 0. • P[S] = 1 if Ei ∩ Ej= f Property 3 is called the additive rule for probability
for any event E Note: hence or Note:
Example –Probabilities for various events in Lotto 6/49 Let E = Event that no money is won
Note: This also could have been calculated by adding up the probability of events where money is won. If it is easier to compute the probability of the compliment of an event, than use the above equation to calculate the probability of the event E.
Example: The birthday problem In a room of n unrelated individuals, what is the probability that at least 2 have the same birthday. Let E = the event that at least 2 have the same birthday. = the event no pair of individuals have the same birthday. n(S) = total number of ways of assigning the n individuals birthdays = 365n = total number of ways of assigning the n distinctbirthdays
Thus Table: 50%
Rule The additive rule P[A B] = P[A] + P[B] – P[A B] and if P[A B] = f P[A B] = P[A] + P[B]
Proof and Ei Ej = f Als0
Hence Example:Saskatoon and Moncton are two of the cities competing for the World university games. (There are also many others). The organizers are narrowing the competition to the final 5 cities.There is a 20% chance that Saskatoon will be amongst the final 5. There is a 35% chance that Moncton will be amongst the final 5 and an 8% chance that both Saskatoon and Moncton will be amongst the final 5. What is the probability that Saskatoon or Moncton will be amongst the final 5.
Hence Solution:Let A = the event that Saskatoon is amongst the final 5.Let B = the event that Moncton is amongst the final 5.Given P[A] = 0.20, P[B] = 0.35, and P[A B] = 0.08What is P[A B]?Note: “and”≡ , “or” ≡ .
Another Example – Bridge Hands A Bridge hand is 13 cards chosen from a deck of 52 cards. Total number of Bridge Hands
Some Questions • What is the probability that the hand contains exactly 5 spades (8 non-spades)? • What is the probability that the hand contains exactly 5 hearts? • What is the probability that the hand contains exactly 5 spades and 5 hearts? • What is the probability that the hand contains exactly 5 spades or 5 hearts? (i. e. a five card major)
Solutions • What is the probability that the hand contains exactly 5 spades (8 non-spades)? Let A = the event that the hand contains exactly 5 spades (8 non-spades)? n(A)= the number of hands that contain exactly 5 spades (8 non-spades)? the number of ways of choosing the 8 non-spades the number of ways of choosing the 5 spades
What is the probability that the hand contains exactly 5 hearts? Let B = the event that the hand contains exactly 5 hearts?
What is the probability that the hand contains exactly 5 spades and 5 hearts? A B = the event that the hand contains exactly 5 spades and 5 hearts? the number of ways of choosing the3 non-spades,hearts the number of ways of choosing the 5 spades the number of ways of choosing the 5 hearts
What is the probability that the hand contains exactly 5 spades or 5 hearts? (i. e. a five card major) Thus there is a 24.26% chance that a bridge hand will contain at least one five card major.
Rule The additive for more than two events For three events P[A B C] = P[A] + P[B] + P[C] – P[A B] – P[A C] – P[B C] + P[A B C] if A B = f, A C = f and B C = f then P[A B C] = P[A] + P[B] + P[C]
B E6 E2 A E3 C E1 E5 E7 E4
A B C = E1 E2 E3 E4 E5 E5 E7 A = E1 E2 E4 E5 When these three are added E2, E3 and E4 are counted twice and E1is counted three times B = E1 E2 E3 E6 C = E1 E3 E4 E7 A B= E1 E2 When these three are subtracted off the extra contributions of E2, E3 and E4 are removed and the contribution of E1is completely removed A C= E1 E4 B C= E1 E3 To correct we now have to add back in the contribution of E1. A B C = E1
An Example Three friends A, B and C live together in the same apartment. For an upcoming “Stanley cup playoff game” there is • An 80% chance that A will watch. • A 60% chance that B will watch. • A 45% chance that C will watch. • A 50% chance that A and B will watch. (and possibly C) • A 30% chance that A and C will watch. (and possibly B) • A 25% chance that B and C will watch. (and possibly A) • A 15% chance that all three (A , B and C) will watch. What is the probability that either A, B or C watch the upcoming “Stanley cup playoff game?”
P[A] = 0.80. • P[B] = 0.60. • P[C] = 0.45. • P[A B] = 0.50. • P[A C] = 0.30. • P[B C] = 0.25. • P[A B C] = 0.15. Thus P[A B C] = P[A] + P[B] + P[C] – P[A B] – P[A C] – P[B C] + P[A B C] = 0.80 + 0.60 + 0.45 – 0.50 – 0.30 – 0.25 + 0.15 = 0.95
Rule The additive rule for more than two events For n events Finally if Ai Aj = f for all i ≠ j. then
Example: This is a Classic problem Suppose that we have a family of n persons. • At Christmas they decide to put their names in a hat. • Each person will randomly pick a name. • This will be the only person for which they will buy a present. Questions • What is the probability that at least one person picks his (or her) own name? • How does this probability change as the size of the family, n, increase to infinity?
Solution: Let Ai = the event that person i picks his own name. We want to calculate: Now because of the additive rule
Note: N = n(S) = the total number of ways you can assign the names to the n people = n! To calculate: we need to calculate:
Now n(Ai) = the number of ways we can assign person i his own name (1) and arbitrarily assign names to the remaining n – 1 persons ((n – 1)!) = 1 (n – 1)! = (n – 1)! : Thus: and
Now n(Ai Aj)= the number of ways we can assign person i and person j their own name (1) and arbitrarily assign names to the remaining n – 2 persons ((n – 2)!) = 1 (n – 2)! = (n – 2)! : Thus: and
Now n(Ai Aj Ak)= the number of ways we can assign person i, person j and person k their own name (1) and arbitrarily assign names to the remaining n – 3 persons ((n – 3)!) = 1 (n – 3)! = (n – 3)! : Thus: and
Continuing n(A1A2 A3… An)= the number of ways we can assign all persons their own name = 1 Thus:
Finally: As the family size increases to infinity ( the number of terms become infinite)
Summary of Rules to date Additive Rules if A and B disjoint if Aiand Ajare all disjoint.
Conditional Probability • Frequently before observing the outcome of a random experiment you are given information regarding the outcome • How should this information be used in prediction of the outcome. • Namely, how should probabilities be adjusted to take into account this information • Usually the information is given in the following form: You are told that the outcome belongs to a given event. (i.e. you are told that a certain event has occurred)
Definition Suppose that we are interested in computing the probability of event A and we have been told event B has occurred. Then the conditional probability of A given B is defined to be:
Rationale: If we’re told that event B has occurred then the sample space is restricted to B. The probability within B has to be normalized, This is achieved by dividing by P[B] The event A can now only occur if the outcome is in of A ∩B. Hence the new probability of A is: A B A ∩ B
An Example The academy awards is soon to be shown. For a specific married couple the probability that the husband watches the show is 80%, the probability that his wife watches the show is 65%, while the probability that they both watch the show is 60%. If the husband is watching the show, what is the probability that his wife is also watching the show
Solution: The academy awards is soon to be shown. Let B = the event that the husband watches the show P[B]= 0.80 Let A = the event that his wife watches the show P[A]= 0.65 and P[A ∩ B]= 0.60
Another Example Suppose a bridge hand (13 cards) is selected from a deck of 52 cards Suppose that the hand contains 5 spades. What is the probability that it also contains 5 hearts. Solution N = n(S) = the total # of bridge hands = Let A = the event that the hand contains 5 hearts Let B = the event that the hand contains 5 spades
Another Example In the dice game – craps, if on the first roll you roll • a 7 or 11 you win • a 2 or 12 you lose, • If you roll any other number {3,4,5,6,8,9,10} that number is your point. You continue to roll until you roll your point (win) or a 7 (lose) Suppose that your point is 5 what is the probability that you win. Repeat the calculation for other values of your point.
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) The Sample Space S N = n(S) = 36
Let B = the event {5} or {7} (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) n(B) = 4 + 6 = 10
Let A = the event {5} = A ∩ B (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) n(A) = n(A ∩ B) = 4
Definition Two events A and B are called independent if Note Thus in the case of independence the conditional probability of an event is not affected by the knowledge of the other event
Difference between independence and mutually exclusive mutually exclusive Two mutually exclusive events are independent only in the special case where Mutually exclusive events are highly dependent otherwise. A and B cannot occur simultaneously. If one event occurs the other event does not occur. A B
