BASIC PROBABILITY. the probability that event A would occur = Pr(A) and. Pr(A) = n A n

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1 BASIC PROBABILITY The classic interpretation... if n = number of mutually exclusive equally likely outcomes (sample space) of which n A constitute event A, then... the probability that event A would occur = Pr(A) and Pr(A) = n A n e.g. suppose that a set of 10 rock samples includes 3 that contain gold nuggets. If you were to pick up a sample at random, what is the probability that it includes a gold nugget? Pr(A) = n A n = 3 10 = 0.3 Addition and multiplication of probabilities given a sequence of independent and mutually exclusive events (E 1, E 2, E 3, E i ), the probability of a given sequence is... addition = "or" Pr (E 1 or E 2 or E 3 or... E i ) = Pr (E 1 )+Pr(E 2 )+Pr(E 3 )... +Pr(E i ) e.g. with one dice, what is the probability of getting a 2 or a 6 in a single role... note n = 6 (six possible outcomes) Pr(2 or 6) = Pr(2)+Pr(6) = 1/6+1/6=1/3 multiplication = "and" Pr(E 1 and E 2 and E 3 and... E i ) = Pr (E 1 ) * Pr(E 2 ) * Pr(E 3 ) *..Pr(E i ) 5.1

2 e.g. with two dice..., what is the probability of getting a 2 and a 6... Pr(2 and 6) = Pr(2) * Pr(6) =1/6 * 1/6 =1/36 = note Pr(2 and 6) = Pr (6 and 2) can mix it up... Pr (E 1 and E 2 or E 3 ) = Pr (E 1 ) * Pr(E 2 +E 3 ) e.g. with two dice... Pr(2 and either 6 or 4) = Pr(2) * Pr(6+4) =1/6 * (1/6+1/6)=2/36 Pr (2 and 6 -or- 6 and 2) = Pr(2)*Pr(6)+Pr(6)*Pr(2) =1/36 + 1/36 = 2/36 Note, the above requires that each roll of the die is independent and mutually exclusive, that is you are rolling two separate die. Joint probability is for two events that are NOT mutually exclusive but are independent. Here we consider (intersection of the two probabilities) and (union of the two probabilities) Pr (A) or Pr(B) = Pr (A) Pr(B) Pr (A B) Conditional probability is for two events in which the outcome for one is dependant on the outcome of the other event. Suppose the probability of event A depends on event B occurring, then

3 Pr(A B) = Pr(A B) Pr(B) Note, if two events (A and B) are independent, this works out to: Pr(A B) = Pr(A and B) Pr(B) or Pr(A) Thus, we use tend conditional probabilities when our events are not independent. Note that Pr(A B) usually doesn t equal Pr(B A), but. Pr(A B) = Pr(B A) Pr(A) Pr(B) This is referred to as Bayes Theorem e.g. suppose the probability of finding species x depends upon the rock type. Given the following data for 120 rock specimens (60 are siltstone and 60 are sandstone), 79 of which contain species x... siltstone sandstone Total samples Species x The chance of finding either a sandstone or siltstone is simply: Pr(sandstone) = Pr(ss) = 60/120, = 0.5 Pr(siltstone) = Pr (slt) = 60/120, =

4 And the probability of finding species x irrespective of rock type? Pr(x) = 79/120 = We can also ask, what is the probability of finding species x if we knew the sample was from a siltstone? Pr(x slt) = Pr(x) Pr(slt) Pr(slt) and Pr(x slt) = Pr(slt x) Pr(x) Pr(x slt) = Pr(x) Pr(slt) Pr(slt) = 57/60 60/120 60/120 =

5 Binomial Distribution (discrete binomial probability) When there are just two possible outcomes, success and failure... p = probability of event occurring (success) q = probability of event not occurring (failure) some examples... p q probability of: heads not heads (tails) probability of: species A not species A probability of: extinction not extinction n = number of trials e.g. the number of coins flipped, number of samples counted, etc... binomial distribution characterized by equation (p+q) n for n=1 p + q for n=2 p 2 + 2pq + q 2 for n=3 p 3 + 3p 2 q + 3pq 2 + q 3 for n=4 p 4 + 4p 3 q + 6p 2 q 2 + 4pq 3 + q 4 deals with combinations not permutations e.g. p * q = q * p to get probabilities, put p and q into the formula for any one term... Pr(x) = n! x!(n x)! p x q n x where x is the number of successes and n is the number of trials, and! is a factorial (e.g. 4!=4*3*2*1 and 3! = 3*2*1) e.g. p = presence of species A = 0.5 q = absence of species A =

6 for n = 1 binomial terms p q probability for n=2 binomial terms p 2 2pq q 2 probability for n=3 binomial terms p 3 3p 2 q 3pq 2 q 3 probability n=1 n=2 n= p 2 q 3pq.25 p q 2pq p q p 3 3 q # of A's # of A's # of A's 2 e.g. presence or absence of trilobite species A p = species A present = q = species A absent = for n = 1 binomial terms p q probability for n=2 binomial terms p 2 2pq q 2 probability for n=3 binomial terms p 3 3p 2 q 3pq 2 q 3 probability

7 n=1 n=2 n= q 2 3p 2 q p q 2pq p q 3pq p Number of A's Some properties of p and q when p = q then distribution is symmetrical when p q then distribution is asymmetrical when n is large and p and q differ greatly (e.g., real small p) becomes Poisson distribution (binomial distribution for rare events). when p = q and n is large then becomes normal distribution if n is large enough, regardless of p and q, the distribution becomes normal (remember Central Limits Theorem). Some other properties of the binomial distribution... x = np (the average number of trials that will contain a success) se = pq n variance = npq standard deviation = npq 5.7

8 Confidence limits of binomial probabilities: p =p ± 1.96 se Testing probabilities For testing a observed probability versus a theoretical one H o : p obs = p theor H o : p obs p theor Use a calculated value Z where Z = p obs p theor se obs compare calculated Z value with critical value (from table) to determine outcome of test For testing the equivalence between two probabilities H o : p A = p Br H o : p A p B Use a calculated value of Z where... Z = p A p B ( se A ) 2 + ( se B ) 2 compare calculated Z value with critical value (from table) to determine outcome of test. 5.8

9 Binomial Frequencies Fitting a theoretical binomial distribution to observed frequencies. e.g. Suppose we collected 1000 m 2 of outcrop (at 1 m 2 samples) and found 401 trilobites. We recorded the number of trilobites per each square meter (0-3). How does our sample distribution fit a binomial distribution? Step 1. determine p, q. Can get total sample probability from simple proportion 401/1000 so... p = and q = Step 2. calculate theoretical binomial probabilities for each x (0-3) by plugging in p and q into equation. Step 3. calculate theoretical frequencies by multiplying through each probability by N Trilobite abundance (x) Observed frequency Binomial term Theoretical probability Pr Theoretical frequency Pr*N q pq p 2 q p test "goodness of-fit" by Chi-square or Kolmogorov-Smirnov test if get good fit between observed and theoretical frequencies (or probabilities), the observed frequencies represent random binomial roll for the observed p, q, and n. The events are independent. 5.9

10 Poisson Distribution Special case of binomial distribution for rare events (p < 0.1) that are widely spaced in time or space ground rules the events are rare 2. the events are independent 3. the probability of an event is constant through the sequences of intervals. Events can be number per time, number per area, number per volume, etc.. 4. probability of an event is proportional to the size of the interval 5. probability of 2 events per interval is much smaller than 1 event per interval 6. whole mess is random role Estimates for Poisson distribution x = np variance =npq usually don't know n or p but can easily calculate x shape of distribution is function of the mean µ=0.1 µ=3 µ=10 Poisson probability Pr(x) = e np (np) x x! = e x x x x! 5.10

11 where x = the number of successes or events per interval and e = the base of natural logs = e.g. collected Ammonites over 200 square meters of Formation x and want to know if are distributed randomly as a Poisson distribution. N = 200 (square meters) x = (Ammonites per square meter) s 2 = e = e x = determine expected Poisson probability by plugging each x (0-5) into the equation. 2 examples are shown below for n = 0: for n = 4: e x x x = = x! 0! e x x x = = x! 4! expected frequency is expected probability times the sample size (N) # Ammonites Collected/m 2 x Observed Frequency xobs Expected Poisson Probability p Expected Frequency xexp can test expected with observed with Chi-square or Kolmogorov- Smirnov test. 5.11

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