The Calculus of Probability



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The Calculus of Probability Let A and B be events in a sample space S. Partition rule: P(A) = P(A B) + P(A B ) Example: Roll a pair of fair dice P(Total of 10) = P(Total of 10 and double) + P(Total of 10 and no double) = 1 36 + 2 36 = 3 36 = 1 12 Complementation rule: P(A ) = 1 P(A) Example: Often useful for events of the type at least one : P(At least one even number) = 1 P(No even number) = 1 9 36 = 3 4 Containment rule P(A) P(B) for all A B Example: Compare two aces with doubles, 1 36 = P(Two aces) P(Doubles) = 6 36 = 1 6 Calculus of Probability, Jan 26, 2003-1 -

The Calculus of Probability Inclusion and exclusion formula P(A B) = P(A) + P(B) P(A B) Example: Roll a pair of fair dice P(Total of 10 or double) = P(Total of 10) + P(Double) P(Total of 10 and double) = 3 36 + 6 36 1 36 = 8 36 = 2 9 The two events are Total of 10 = {46, 55, 64} and Double = {11,22,33,44,55,66} The intersection is Total of 10 and double = {55}. Adding the probabilities for the two events, the probability for the event 55 is added twice. Calculus of Probability, Jan 26, 2003-2 -

Conditional Probability Probability gives chances for events in sample space S. Often: Have partial information about event of interest. Example: Number of Deaths in the U.S. in 1996 Cause All ages 1-4 5-14 15-24 25-44 45-64 65 Heart 733,125 207 341 920 16,261 102,510 612,886 Cancer 544,161 440 1,035 1,642 22,147 132,805 386,092 HIV 32,003 149 174 420 22,795 8,443 22 Accidents 1 92,998 2,155 3,521 13,872 26,554 16,332 30,564 Homicide 2 24,486 395 513 6,548 9,261 7,717 52 All causes 2,171,935 5,947 8,465 32,699 148,904 380,396 1,717,218 1 Accidents and adverse effects, 2 Homicide and legal intervention measure probability with respect to a subset of S Conditional probability of A given B P(A B) = P(A B), if P(B) > 0 P(B) If P(B) = 0 then P(A B) is undefined. Conditional probabilities for causes of death: P(accident) = 0.04282 P(age=10) = 0.00390 P(accident age=10) = 0.42423 P(accident age=40) = 0.17832 Calculus of Probability, Jan 26, 2003-3 -

Conditional Probability Example: Select two cards from 32 cards What is the probability that the second card is an ace? P(2nd card is an ace) = 1 8 What is the probability that the second card is an ace if the first was an ace? P(2nd card is an ace 1st card was an ace) = 3 31 Calculus of Probability, Jan 26, 2003-4 -

Multiplication rules Example: Death Rates (per 100,000 people) All Ages 1-4 5-14 15-24 25-44 45-64 65 872.5 38.3 22.0 90.3 177.8 708.0 5071.4 Can we combine these rates with the table on causes of death? What is the probability to die from an accident (HIV)? What is the probability to die from an accident at age 10 (40)? Know P(accident die) = P(die from accident)/p(die) P(die from accident) = P(accident die)p(die) Calculate probabilities: P(die from accident) = 0.04281 0.00873 = 0.00037 P(die from accident age = 10) = 0.42423 0.00090 = 0.00038 P(die from accident age = 40) = 0.17832 0.00178 = 0.00031 P(die from HIV) = 0.01473 0.00873 = 0.00013 P(die from HIV age = 10) = 0.02055 0.00090 = 0.00002 P(die from HIV age = 40) = 0.15308 0.00178 = 0.00027 General multiplication rule P(A B) = P(A B)P(B) = P(B A)P(A) Calculus of Probability, Jan 26, 2003-5 -

Independence Example: Roll two dice What ist the probability that the second die shows 1? P(2nd die = 1) = 1 6 What ist the probability that the second die shows 1 if the first die already shows 1? P(2nd die = 1 1st die = 1) = 1 6 What ist the probability that the second die shows 1 if the first does not show 1? P(2nd die = 1 1st die 1) = 1 6 The chances of getting 1 with the second die are the same, no matter what the first die shows. Such events are called independent: The event A is independent of the event B if its chances are not affected by the occurrence of B, P(A B) = P(A). Equivalently, A and B are independent if P(A B) = P(A)P(B) Otherwise we say A and B are dependent. Calculus of Probability, Jan 26, 2003-6 -

Let s Make a Deal The Rules: Three doors - one price, two blanks Candidate selects one door Showmaster reveals one loosing door Candidate may switch doors 1 2 3 Would YOU change? Can probability theory help you? What is the probability of winning if candidate switches doors? What is the probability of winning if candidate does not switch doors? Calculus of Probability, Jan 26, 2003-7 -

The Rule of Total Probability Events of interest: A - choose winning door at the beginning W - win the price Strategy: Switch doors (S) Know: PS(W A) = 0 PS(W A ) = 1 PS(A) = 1 3 PS(A ) = 2 3 Probability of interest: PS(W ): P S (W ) = PS(W A) + PS(W A ) = PS(W A)PS(A) + PS(W A )PS(A ) = 0 1 3 + 1 2 3 = 2 3 Strategy: Do not switch doors (N) Know: PN(W A) = 1 PN(W A ) = 0 PN(A) = 1 3 PN(A ) = 2 3 Probability of interest: PN(W ): P N (W ) = PN(W A) + PN(W A ) = PN(W A)PN(A) + PN(W A )PN(A ) = 1 1 3 + 0 2 3 = 1 3 Calculus of Probability, Jan 26, 2003-8 -

The Rule of Total Probability Rule of Total Probability If B 1,..., B k mutually exclusive and B 1... B k = S, then P(A) = P(A B 1 )P(B 1 ) +... + P(A B k )P(B k ) Example: Suppose an applicant for a job has been invited for an interview. The chance that he is nervous is P(N) = 0.7, the interview is succussful if he is nervous is P(S N) = 0.2, the interview is succussful if he is not nervous is P(S N ) = 0.9. What is the probability that the interview is successful? P(S) = P(S N)P(N) + P(S N )P(N ) = 0.2 0.7 + 0.9 0.3 = 0.441 Calculus of Probability, Jan 26, 2003-9 -

The Rule of Total Probability Example: Suppose we have two unfair coins: Coin 1 comes up heads with probability 0.8 Coin 2 comes up heads with probability 0.35 Choose a coin at random and flip it. What is the probability of its being a head? Events: H= heads comes up, C 1 = 1st coin, C 2 = 2nd coin P(H) = P(H C 1 )P(C 1 ) + P(H C 2 )P(C 2 ) = 1 (0.8 + 0.35) = 0.575 2 Calculus of Probability, Jan 26, 2003-10 -

Example: O.J. Simpson Bayes Theorem Only about 1 10 of one percent of wife-batterers actually murder their wives Lawyer of O.J. Simpson on TV Fact: Simpson pleaded no contest to beating his wife in 1988. So he murdered his wife with probability 0.001? Sample space S - married couples in U.S. in which the husband beat his wife in 1988 Event H - all couples in S in which the husband has since murdered his wife Event M - all couples in S in which the wife has been murdered since 1988 We have P(H) = 0.001 Then P(M H) = 1 since H M P(M H ) = 0.0001 at most in the U.S. P(H M) = P(M H)P(H) P(M) P(M H)P(H) = P(M H)P(H) + P(M H )P(H ) 0.001 = 0.001 + 0.0001 0.999 = 0.91 Calculus of Probability, Jan 26, 2003-11 -

Bayes Theorem Reversal of conditioning (general multiplication rule) P(B A)P(A) = P(A B)P(B) Rewriting P(A) using the rule of total probability we obtain Bayes Theorem P(B A) = P(A B)P(B) P(A B)P(B) + P(A B )P(B ) If B 1,..., B k mutually exclusive and B 1... B k = S, then P(B i A) = P(A B i )P(B i ) P(A B 1 )P(B 1 ) +... + P(A B k )P(B k ) (General form of Bayes Theorem) Calculus of Probability, Jan 26, 2003-12 -

Bayes Theorem Example: Testing for AIDS Enzyme immunoassay test for HIV: P(T+ I+) = 0.98 (sensitivity - positive for infected) P(T- I-) = 0.995 (specificity - negative for noninfected) P(I+) = 0.0003 (prevalence) What is the probability that the tested person is infected if the test was positive? P(I+ T+) = = P(T+ I+)P(I+) P(T+ I+)P(I+) + P(T+ I-)P(I-) 0.98 0.0003 0.98 0.0003 + 0.005 0.9997 = 0.05556 Consider different population with P(I+) = 0.1 (greater risk) P(I+ T+) = 0.98 0.1 0.98 0.1 + 0.005 0.9 = 0.956 testing on large scale not sensible (too many false positives) Repeat test (Bayesian updating): P(I+ T++) = 0.92 in 1st population P(I+ T++) = 0.9998 in 2nd population Calculus of Probability, Jan 26, 2003-13 -