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Chapter 2
Julian Chan
May 7, 2012
1 More about sets and the rules of probability
Recall that for any event A ⊂ S we have:
0 ≤ P (A) ≤ S, and P (S) = 1.
There are several other axioms of probability that will be important. The first new one is that Ai are mutually exclusive events then
P (∪ii==1n Ai) =
i∑=n
i=
Ai
Example 1. Consider the experiment of tossing two dice. In this instance S = HH, HT, T H, T T , and we define A1 = HH, HT , A2 = T H. It is easy to see that
P (S) = 1 and P (A 1 ∪ A 2 ) =
= P (A1) + P (A2).
Another immediate consequence is that
P (A) = 1 − P (Ac)
Example. From an ordinary deck of 52 cards, seven cards are drawn at random and without replacement. What is the probability that at least one of the cards is a king? You can get 7 cards in
7 waysandyoucanchooseanycardotherthanakingin
7
ways therefore
P (at least one king) = 1 − P (no kings) = 1 −
7
7
2 General addition rule
Example 2. Consider the experiment of tossing two dice. In this instance S = HH, HT, T H, T T , and we define A1 = HH, HT , A2 = HT, T H, B = T H. Notice that
P (A 1 ∪ B) =
= P (A1) + P (B) and P (A 1 ∪ A 2 ) =
= P (A1) + P (A2) = 1.
The question is why can we add ∪ (OR) probabilities only in some instances? What’s going on man? Notice that in the first instance A 1 ∪B = 0 while in the second A 1 ∪A2 = HT 6 = 0. The probability P (A1) + P (A2) is over counting by exactly P (A 1 ∩ A2) = 14! We thus have,
P (A 1 ∪ A 2 ) =
= P (A1) + P (A2) − P (A 1 ∩ A2) = 1 −
This leads us to the general addition rule which is:
P (A 1 ∪ A 2 ) = P (A1) + P (A2) − P (A 1 ∩ A2).
Can you draw the VENN DIAGRAM???
Example. Experience shows that 25% of all complaints about the telephone lines involve satic, and 50% involve line deterioration. 35% involve only line deterioration. What is the probability that a randomly selected complaint will involve both A =static and B = line deterioration? What about neither problem? First you should draw the VENN DIAGRAM! Now identify the given information:
P (A) = 25% P (B) = 50% P (B ∩ Ac) = 35%.
We want to find P (A ∩ B). We can see with the use of a venn diagram P (A ∩ B) = P (B) − P (B ∩ Ac) =. 5 − .35 =. 15 newline The probability of neither is given by
P (neither) = 1 − P (A ∪ B) = 1 − (P (A) + P (B) − P (A ∩ B)) = 1 − (.25 +. 5 − .15) =. 4.
3 Conditional probability
Conditionaly probability describes the situation in whcih an experiment will occur, but you know that only certain outcomes are possible or not possible.
Example. Consider the experiment of rolling two dice. Let event A be the sum of the face values is 5 , and B the event that die one is a 7. We know from the general counting principle that the total number of possible outcomes is 36. We can also verify that:
A = (1, 4), (4, 1), (2, 3), (3, 2) and B = (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6).
From this we calculate that
P (A) =
P (B) =
and P (A ∩ B) =
Suppose that you know that event A will occur! Since A must occur this changes the SAMPLE SPACE to
S = A = (1, 4), (4, 1), (2, 3), (3, 2)
probability 30% if John is innocent. What is the probability that in the course of the trial that Greg and Eva will give conflicting testimony? Let I be the event John is innocient and C the event that Greg and Eva give conflicting testimony.
P (C) = P ((C ∩ I) ∪ (C ∩ Ic)) = P ((C ∩ I)) + P ((C ∩ Ic))
= P ((C|I) × P (I) + P ((C|Ic) × P (Ic) =. 3 × .35 +. 25 × .65 =. 26
4 Independce
The notion of indepence is that the occurance of one event does not change the odds of another event. Naturally there are many examples in which two events are dependent such as knowing that the next card drawn will be a king will affects the odds of a poker player. The notion of independce can also be expressed in terms of conditional probabilities. The idea is that since knowing event A will not change the probability of B it is reasonable to say that A and B are independent IFF:
P (B ∩ A) = P (A) × P (B).
Consequently we have that if A and B are independent then
P (B|A) =
P (A ∩ B)
P (B)
P (A) × P (B)
P (B)
= P (A).
Example. Let us draw a card from a deck of 52 cards, A is the event that the card is a 5 or 6 , and B is the event that the card is heart. Are the events A and B independent?
P (A ∩ B) =
= P (A) × P (B) =
×
This computation tells us that the events A and B are independent even tough they have non empty intersection! QUESTION: if two events have empty intersection are they independent or not?
Example. (JAILER’s PARADOX) The jailer of a prision in which Alex, Bill, and Tim are held is the only person, other than the judge, who knows which of these three prisioners is condemned to death, and which two will be freed. Alex has written a letter to his unfortunate fiancee and wants to give it to either Tim or Bill, whomever goes free, to deliver. Alex asks the jailer to tell him which of the two will be freed. The jailer refuses to give the information explaining that if he does, the probability of Alex dying increases from 13 to 12. This seems intutively suspect, since the jailer is providing no new information to Alex so why should the probability of dying change? Alex already knows that one of the other two prisioners will go free and obviously, knowing the name of the one who will be freed does not change the probability of his dying. To explain this paradox Alex will show even if the jailer tells the name of which of the other two will go free the
probability of his dying is still 13. Let the events A, B, and T represent the events Alex dies, Bill dies, and Tim dies respectively. Let C1 = (T, the jailer tells Alex that Bill goes free), C 2 = (B, the jailer tells Alex that Tim goes free), C3 = (A, the jailer tells Alex that Bill goes free), and C4 = (A, the jailer tells Alex that Tim goes free). The sample space is S = C 1 , C 2 , C 3 , C 4. We assume that P (C1) = 13 = P (C2). To assign probabilites to C 3 and C 4 we assume the event A is independent of the jailers statment to Alex, and if Alex is the one scheduled to die then it is fifty fifty if he says Tim or Bill goes free hence P (A3) = P (A4) = 16. Let J be the event taht the jailer tells Alex tim goes free; then
P (A|J) =
P (A ∩ J)
P (J)
P (C4)
P (C2) + P (C4)
1 6 1 3 +^
1 6
so showing that even if the jailer reveals which of the other two who will go free, the probability of Alex dying is still 13.
5 Bayes formula
Bayes formula can be very useful for computing probabilities involving conditional state- ments. The formula is:
Theorem. Let A 1 ,... An be a collection of mutually exclusive events whose untions is S, and B is an event with positive probability then
P (Aj |B) = ∑P^ (B|Aj^ )^ ×^ P^ (Aj^ ) n i=1 P^ (B|Ai^ ×^ P^ (Ai)
Example. A judge is 65% sure that Mary has committed a crime. During the course of the trial a witness convinces the judge that there is an 85 chance that the crimal is left handed. If 23% of the population is left handed, and Mary is also left handed; with this information how certain should the judge be of the guilt of Mary? Let G and I denote the events of guilty, and innocent, respectively. Let A be the event Mary is left handed, Bayes formula gives
P (Gj |A) =
P (A|G) × P (G)
P (A|I) × P (I) + P (A|G) × P (G)
. 85 ×. 65
. 23 × .35 +. 85 ×. 65
