The Binomial distribution

Examples and Definition

Binomial Model (an experiment ) 1 A series of n independent trials is conducted. 2 Each trial results in a binary outcome (one is labeled success the other failure ). 3 The probability of success is equal to p for each trial, regardless of the outcomes of the other trials.

Binomial Random Variable The number of successes in the binomial experiment. Let Y = # of success in the above model. Then Y is a binomial random variable with parameters n (sample size) and p (success probability). It is often denoted Y B(n, p).

Example: Tossing a fair coin Toss a fair coin three times, P(H) =.5. Interest is counting the number of heads. Y = # of heads success = heads; failure = tails n = p = Y

Example: Tossing a fair coin Toss a fair coin three times, P(H) =.5. Interest is counting the number of heads. Y = # of heads success = heads; failure = tails n = 3 p = Y

Example: Tossing a fair coin Toss a fair coin three times, P(H) =.5. Interest is counting the number of heads. Y = # of heads success = heads; failure = tails n = 3 p =.5 Y

Example: Tossing a fair coin Toss a fair coin three times, P(H) =.5. Interest is counting the number of heads. Y = # of heads success = heads; failure = tails n = 3 p =.5 Y B(3,.5)

Example: Tossing an unfair coin Toss a biased coin 5 times, P(H) =.7. Interest is counting the number of heads. Y = # of heads success = heads; failure = tails n = 5 p =.7 Y B(5,.7)

Example: Counting Mutations Experiment to mutate a gene in bacteria; the probability of causing a mutation is.4. The experiment was repeated 10 times, with 10 independent colonies. Interest is counting the number of mutations. Y = # of mutations. success = mutation; failure = no mutation n = 10 p =.4 Y B(10,.4)

Computing Probabilities for a Binomial Random Variable.

Board Example Tossing a biased coin; Y B(3,.7).

Tossing a fair Coin Consider tossing a fair coin 3 times. Y = # of heads in the 3 tosses; Y B(3,.5) Consider the Possible outcomes: TTT HTT THT TTH HHT HTH THH HHH IP{Y = 0} = 1 2.1 2.1 2 = 1 ( 8 ) 1 IP{Y = 1} = 3 2.1 2.1 = 3 2 8 ( ) 1 IP{Y = 2} = 3 2.1 2.1 2 ) = 3 8 IP{Y = 3} = 1 2.1 2.1 2 = 1 8

Tossing a Biased Coin Consider tossing a biased coin 3 times; IP{H} =.7 Y = # of heads in the 3 tosses; Y B(3,.7) Consider the Possible outcomes: TTT HTT THT TTH HHT HTH THH HHH IP{Y = 0} = 1(.3.3.3) = 1(.7 0 )(.3 3 ) =.027 IP{Y = 1} = IP{Y = 2} = IP{Y = 3} =

Question Is there a general formula for computing Binomial probabilities? We do not want to have to list all possibilities when n = 10, or n = 100.

Background: Factorials Factorial. Multiply all numbers from 1 to n (n is a positive integer) n! = n(n 1)(n 2)...(2)(1) Example. n = 4. Note. 0! = 1. n! = (4)(3)(2)(1) = 24.

Background: Binomial Coefficients nc j nc j = n! j!(n j)! nc j counts all of the ways that j successes can appear in n total trials. Example n = 5, j = 3. 5! 3!2! = (5)(4) = 10. 2 cf. Table 2 p.674 for the n C j numbers.

Chalkboard Y B(n, p) Can we guess the answer: IP{Y = j} =?

Binomial Distribution Formula IP{Y = j} = n C j p j (1 p) n j Y B(n, p) (a binomial random variable based on n trials and success probability p) The probability of getting j successes is given by (j = 0, 1,..., n) IP{Y = j} = n C j p j (1 p) n j = n! j!(n j)! pj (1 p) n j

Example IP{Y = j} = n! j!(n j)! pj (1 p) n j Question: You toss a fair coin 3 times, what is the probability of getting 2 heads. Answer: p =.5 (fair coin), n = 3 tosses, j = 2 Heads. We get IP{Y = 2} = 3! 2! 1! (.5)2 (.5) 1 = 3.2.1 2.1 1 (.5)3 = 3/8

Example IP{Y = j} = n! j!(n j)! pj (1 p) n j Question: Y B(7,.6). Compute IP{Y = 2}. Answer: p =.6, n = 7 tosses, j = 2 Heads. We get IP{Y = 2} = 7! 2! 5! (.6)2 (.4) 5 7.6 5! = 1.2 5! (.6)2 (.4) 5 = 21(.6) 2 (.4) 5 =.0774

Example A new drug is available. Its success rate is 1/6: probability that a patient is improved. I try it independently on 6 patients. Probability that at least one patient improves? p = 1/6, n = 6, j = 1, 2, 3, 4, 5 or 6. IP{at least one improves} = IP{Y = 1 or Y = 2 or... or Y = 6} = IP{Y = 1} + IP{Y = 2} + + IP{Y = 6} = 1 IP{Y = 0} = 1 6! 0!6! (1/6)0 (5/6) 6 = 1 (5/6) 6 =.665

Probability Distribution IP{Y = j} = n! j!(n j)! pj (1 p) n j Y B(6, 1/6); n = 6, p = 1/6. We can compute IP{Y = j}, for j = 0, 1, 2, 3, 4, 5, 6. y 0 1 2 3 4 5 6 IP{Y = y} 0.335 0.402 0.200 0.054 0.008 0.0006 0.00002

Mean, Variance, and Standard Deviation for Binomial Random Variables

Recall the Formula for Mean,Variance for a General Discrete Random Variable E(Y ) = y i IP{Y = y i }, Var(Y ) = (y i µ Y ) 2 IP{Y = y i }, where the y i s are the values that the variable takes on and the sum is taken over all possible values. What if Y Bin(100,.5) Do I have to sum over 101 different values?!?

Mean, Variance, and Standard deviation If Y B(n, p) then µ = IEY = np, σ 2 = np(1 p), and σ = np(1 p)

Example Coin tossing: Y B(100,.5). Compute the mean, variance, and standard deviation. µ Y = np = 50 σ 2 Y = np(1 p) = 100(.5)(.5) = 25 σ Y = 25 = 5.

The assumptions for the binomial model

Underlying assumptions A binomial random variable satisfies the following four conditions, abbreviated BInS. 1 Binary outcomes. There are two possible outcomes for each trial (success and failure). 2 Independent trials. The outcomes of the trials are independent of each other. 3 n is fixed. The number of trials n is fixed in advance. 4 Same value of p. The probability of a success on a single trial is the same for all trials. Note The binomial model with n trials is said to be made up of n Bernoulli trials.

Bernoulli Trials The Assumptions of Bernoulli Trials. 1 Each trial results in one of two possible outcomes, denoted success (S ) or failure (F ). 2 The trials are independent. 3 The probability of S remains constant from trial-to-trial and is denoted by p. Write q = 1 p for the constant probability of F.

Example - solve on the Chalk Board IP{Y = j} = n! j!(n j)! pj (1 p) n j Question: Devin Harris likes to play basketball. Assume that Devin s free throw attempts are Bernoulli trials with p =.8. 1 Devin will shoot four free throws. Calculate the probability that he obtains S,S,F,S in that order 2 Devin will shoot four free throws. Calculate the probability that he obtains a total of three successes. 3 Can you think of any reasons why the assumptions of the binomial model might not be satisfied for free throw shooting?

Examples for You Example 3.45, p.106 Examples 3.47, 3.48 p.108-109 Example 3.49 p.109 Example 3.50 p.110 (when assumptions might be violated).