More Discussion of the Binomial Distribution: Comments & Examples j Thermo & Stat Mech - Spring 2006 Class 16

Requirements justifying use of the Binomial Distribution: The Binomial Distribution applies ONLY to cases where there are only 2 possible outcomes: heads or tails, success or failure, defective or good item, etc. Requirements justifying use of the Binomial Distribution: 1. The experiment must consist of n identical trials. 2. Each trial must result in only one of 2 possible outcomes. 3. The outcomes of the trials must be statistically independent. 4. All trials must have the same probability for a particular outcome.

Binomial Distribution 3/21/06 Binomial Distribution The Probability of n Successes out of N Attempts is: p = Probability of a Success q = Probability of a Failure q = 1 – p, (p + q)N = 1

Mean of the Binomial Distribution 3/21/06 Mean of the Binomial Distribution l Thermo & Stat Mech - Spring 2006 Class 16

Standard Deviation (s) of the Binomial Distribution 3/21/06 Standard Deviation (s) of the Binomial Distribution 2 1  3

For The Binomial Distribution 3/21/06 For The Binomial Distribution

Common Notation for the Binomial Distribution r items of one type & (n – r) of a second type can be arranged in nCr ways. Here: ≡ nCr is called the binomial coefficient In this notation, the probability distribution is: Wn(r) = nCrpr(1-p)n-r ≡ probability of finding r items of one type & n – r items of the other type. p = probability of a given item being of one type .

Binomial Distribution: Example Problem: A sample of n = 11 electric bulbs is drawn every day from those manufactured at a plant. The probabilities of getting defective bulbs are random and independent of previous results. Probability that a given bulb is defective is p = 0.04. 1. What is the probability of finding exactly three defective bulbs in a sample? (Probability that r = 3?) 2. What is the probability of finding three or more defective bulbs in a sample? (Probability that r ≥ 3?)

Binomial Distribution, n = 11 Number of Defective Bulbs, r Probability 11Crpr(1-p)n-r p = 0.04 11C0 (0.04)0(0.96)11 = 0.6382 1 11C1 (0.04)1(0.96)10 = 0.2925 2 11C2 (0.04)2(0.96)9 = 0.0609 3 11C3 (0.04)3(0.96)8 = 0.0076 l Thermo & Stat Mech - Spring 2006 Class 16

P(r = 3 defective bulbs) = Question 1: Probability of finding exactly three defective bulbs in a sample? P(r = 3 defective bulbs) = W11(r = 3) = 0.0076 l Thermo & Stat Mech - Spring 2006 Class 16

P(r = 3 defective bulbs) = Question 1: Probability of finding exactly three defective bulbs in a sample? P(r = 3 defective bulbs) = W11(r = 3) = 0.0076 Question 2: Probability of finding three or more defective bulbs in a sample? P(r ≥ 3 defective bulbs) = 1- W11(r = 0) – W11(r = 1) – W11(r = 2) = 1 – 0.6382 - 0.2925 – 0.0609 = 0.0084 l Thermo & Stat Mech - Spring 2006 Class 16

Binomial Distribution, Same Problem, Larger r Number of Defective Bulbs, r Probability 11Crpr(1-p)n-r 11C0(0.04)0(0.96)11 = 0.638239 1 11C1 (0.04)1(0.96)10 = 0.292526 2 11C2 (0.04)2(0.96)9 = 0.060943 3 11C3 (0.04)3(0.96)8 = 0.007618 4 11C4 (0.04)4(0.96)7 = 0.000635 5 11C5 (0.04)5(0.96)6 = 0.000037 Thermo & Stat Mech - Spring 2006 Class 16 l

Thermo & Stat Mech - Spring 2006 Class 16 Binomial Distribution n = 11, p = 0.04 Distribution of Defective Items Distribution of Good Items l Thermo & Stat Mech - Spring 2006 Class 16

The Coin Flipping Problem 3/21/06 The Coin Flipping Problem Consider a perfect coin. There are only 2 sides, so the probability associated with coin flipping is The Binomial Distribution. Problem: 6 perfect coins are flipped. What is the probability that they land with n heads & 1 – n tails? Of course, this only makes sense if 0 ≤ n ≤ 6! For this case, the Binomial Distribution has the form: l Thermo & Stat Mech - Spring 2006 Class 16

Binomial Distribution for Flipping 1000 Coins 3/21/06 Binomial Distribution for Flipping 1000 Coins Note: The distribution peaks around n = 500 successes (heads), as we would expect n = Np = 500 l Thermo & Stat Mech - Spring 2006 Class 16

Binomial Distribution for Selected Values of n & p Thermo & Stat Mech - Spring 2006 Class 16

Binomial Distribution for Selected Values of n & p n = 5, p = 0.1 n = 5, p = 0.5 n = 10, p = 0.5 l Thermo & Stat Mech - Spring 2006 Class 16

Binomial Distribution for Selected Values of n & p Thermo & Stat Mech - Spring 2006 Class 16

Binomial Distribution for Selected Values of n & p