9/6/2016. Section 5.1 Probability. Equally Likely Model. The Division Rule: P(A)=#(A)/#(S) Some Popular Randomizers.

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1 Chapter 5: Probability and Discrete Probability Distribution Learn. Probability Binomial Distribution Poisson Distribution Some Popular Randomizers Rolling dice Spinning a wheel Flipping a coin Drawing cards Basic Concepts Outcome: possible result Sample Space(S): all possible outcomes Event(A): a subset of sample space Complement event(a ): All outcomes not in A Mutually exclusive events: Two events can not occur simultaneously Section 5.1 Probability Equally Likely Model The Division Rule: P(A)=#(A)/#(S) More concepts Roll a Die What is the probability of rolling a 6? a..22 b..10 c..17 What is the probability of not rolling a 6? a. 1/6 b. 3/6 c. 5/6 The Subtraction Rule: P(A )=1-P(A) Example use: P(at least one)=1-p(none) Independent events: the outcome of any one trial is not affected by the outcome of any other trial: P(A and B)=P(A)P(B) Example: Roll 5 dice. What is the probability of getting at least one 6? Solution: The complement event of at least one is none. P(no 6 in rolling 5 dice)=(5/6)(5/6)(5/6)(5/6)(5/6) P(at least one 6 in rolling 5 dice) =1-P(no 6) =1-5 5 /6 5 =1-3125/7776=4651/7661 1

2 Example If a family has four girls in a row and is expecting another child, does the next child have more than a ½ chance of being a boy? Probability Distribution A listing, in the form of a table or graph, of the probabilities associated with each possible outcomes Example: outcome probability girl 1/2 boy 1/2 Roll a dice: (1,1/6), (2,1/6),,(6,1/6) 5.2 Counting Techniques Permutation: The number of different sequences considering order Choose r from n objects considering order npr=n(n-1) (n-r+1)=n!/(n-r)! 3P2=3(2)=6, 5P3=5(4)(3)=60 Factorial npn=n! 3P3=3(2)(1)=6 Combination: The number of difference ways without considering order Choose r from n objects without considering order ncr=n!/((n-r)!r!)-binomial coefficient 3C2=3!/(1!2!)=3 5.3 More Probability Rules Substraction Rule: P(A )=1-P(A) Multiplication Rule: If A and B are independent, then P(A and B)=P(A)P(B) Conditional Probability The probability of a second event (B) occurring depends on whether or not the first event (A) occurred. Denoted by P(B A), read as the probability of B given A. Example: Children: A={1 st child is a girl}, B={2 nd child is a girl} P(A)=1/2 P(B)=1/2=P(B A), A and B are independent P(A and B)=P(both are girls)=(1/2)(1/2)=1/4. 2

3 Example More Probability Rules The Multiplication Rule (the and rule) Let A and B be two independent events P(A and B)=P(A) * P(B) In general, P(A and B)=P(A)*P(B A)=P(B)*P(A B) The Addition Rule (the or rule) Let A and B be two mutually exclusive (not overlap) events, then P(A or B)=P(A)+P(B) In general, P(A or B)=P(A)+P(B)-P(A and B) In gambling, you are dealing a deck of 52 cards. If you were to deal 2 cards without replacement, what is the probability of getting 2 kings? Solution: Use tree diagram and multiplication rule P(1 st king)=4/52, P(2 nd king 1 st king)=3/51 Thus, P(2 kings)=(4/52)(3/51)=(1/13)(1/17)=1/221. Denote A={1 st king}, B={2 nd king}. Question: Are A and B independent? Example 1 (continued) What is the probability of getting a king followed by a queen? Solution: P(1 st king)=4/52=1/13 P(2 nd queen 1 st king)=4/51 P(a king followed by a queen) =(1/13)(4/51)=4/663. Example (continued) A={king}, P(A)=4/52 B={heart}, P(B)=13/52 P(A and B)=1/52 P(A or B)=P(A)+P(B)-P(A and B) =4/52+13/52-1/52=16/52=4/ Binomial Distribution Bernoulli trial Independent Bernoulli Trials Example: Toss coins (Punnett square treatment in genetics) Coins are independent Toss two coins: HH, HT, TH and TT Coin 1 H (p=.5) T(p=.5) Coin 2 H (p=.5) HH (.25) TH (.25) p=.5 and q=1-p=.5 T (p=.5) TH (.25) TT(.25) The probabilities for all four outcomes are p 2,pq, qp, q 2 and (p 2 +pq+qp+q 2 )=(p+q) 2 =1. 3

Analogy of Tossing Coins Toss 3 coins and p=p(h) and q=p(t) Possible outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT Associated probabilities are p 3, p 2 q,p 2 q, p 2 q, pq 2,

Then the distribution of X is called Binomial distribution with probabilities given below: n! p ( x) k!( n k)!

9722 For each litter, there are only two possibilities: consists (5 or more males) or not. It is a Bernoulli trial.

4 Analogy of Tossing Coins Toss 3 coins and p=p(h) and q=p(t) Possible outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT Associated probabilities are p 3, p 2 q,p 2 q, p 2 q, pq 2, pq 2, pq 2, q 3 and p 3 +3p 2 q+3pq 2 +q 3 =(p+q) 3 =1 Binomial Distribution (Toss k coins) Let p=p(h)=p(success) X=the number of Heads in k tosses X takes values 0,1,2,,k. Then the distribution of X is called Binomial distribution with probabilities given below: n! p ( x) k!( n k)! p x q k x Bar chart Cumulative Probability Summary Parameters Toss two dice. X=the number of ones~bin(2, 1/6). x P(x) P(Xx) For each litter, there are only two possibilities: consists (5 or more males) or not. It is a Bernoulli trial. Litter to Litter are assumed to be independent Let X= the number of litters consisting 5 or more males in the 100 litters. Then X is Binomial distributed with n=100 and p(5 or more)=.1093=p Total Expectation of X: =E(X) =np=10.93 Variance of X: 2 = Var(X)=np(1-p) Var(X)=np(1-p)=10.93*( )= Standard deviation: =9.7354=

Application to Biological Science Example 5.3: Bush hogs Litters of 6 hoglets P(male)=P(female)=.5 Question: How many litters in n=100 litters will consist of 5 or more males?

0937+.0156=.1093 Step 2: Expect to have np=100*.1093=10.93 litters consisting 5 or more males. Application Males #(x) Females # P(x) Expected Frequnecy p(x)*100 0 6 0.0156 1.56 1 5 0.0937 9.37 2 4 0.

5 Poisson Distribution To model probability of rare event within an interval of time or space Such events are assumed independent Poisson random variable X=the number of occurrences of the rare event

5 Application to Biological Science Example 5.3: Bush hogs Litters of 6 hoglets P(male)=P(female)=.5 Question: How many litters in n=100 litters will consist of 5 or more males? Solution: Notice that each hoglet is either male or female. Step 1: Assuming the six hoglets in a litter are independent being male or female. For each litter, compute p=p(5 or more males)=p(5)+p(6)= =.1093 Step 2: Expect to have np=100*.1093=10.93 litters consisting 5 or more males. Application Males #(x) Females # P(x) Expected Frequnecy p(x)* Poisson Distribution To model probability of rare event within an interval of time or space Such events are assumed independent Poisson random variable X=the number of occurrences of the rare event in a unit time interval. Example: Spatial distribution of plants Poisson Distribution X takes values on {0,1, 2, } The associated probability at x is x e p( x) x! where is estimated by x x e x! µ=the population mean occurrences of the event per sampling unit x = the sample mean. x Example: Maple seedlings µ is estimated by the sample mean which is The probability of 1 seedling/quadrat is estimated by p(1) P( X 1) e ! Compare observed with Poisson The expected is computed based on Poisson distribution if the maple seedlings were distributed randomly within the sampled habitat. 5

Expected Observed 9/6/2016 Poisson distribution in Ecology Test for independence, hence distributed randomly in space or time. Figure 5.7 shows not match well indicating possible not independent.

000 indicating not Poisson distributed Value 40 30 20 10 Chart of Observed and Expected Values 0 Category 0 1 2 3 4 5 6 Historical Test Contribution Observed Counts Proportion Expected Category to

6 Expected Observed 9/6/2016 Poisson distribution in Ecology Test for independence, hence distributed randomly in space or time. Figure 5.7 shows not match well indicating possible not independent. Stat>Tables>Chi-square Goodness-of-fit test gives p-value=0.000 indicating not Poisson distributed Value Chart of Observed and Expected Values 0 Category Historical Test Contribution Observed Counts Proportion Expected Category to Chi-Sq Poisson Distribution applied to a microbial genetics problem Homework exercises 5.1, 5.3, , , 5.13, 5.15, 5.17, ,

Conditional Probability

Conditional Probability Idea have performed a chance experiment but don t know the outcome (ω), but have some partial information (event A) about ω. Question: given this partial information what s the