Binomial Distribution Each slide has its own narration in an audio file. For the explanation of any slide click on the audio icon to start it. Professor Friedman's Statistics Course by H & L Friedman is licensed under a  Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 

Permutations & Combinations First a slight digression to review permutations and combinations. Permutations. A permutation is a particular arrangement.   Example: How many ways can you arrange the letters A, B, and C? ABC, ACB, BAC, BCA, CAB, CBA. Answer:  6 Binomial Distribution

Permutations & Combinations How did we get this? Let’s use 3 imaginary slots. In slot one, we have three letters to choose from, A, B, or C. In slot two, we only have two letters left. Finally, in slot three, we only have one unused letter. Binomial Distribution

Permutations & Combinations On your scientific calculator, you will use the nPr key. The P stands for permutations. n is the number of distinct objects you wish to arrange and r is the number of slots or spaces. Thus, in the above example, we have 3 objects to arrange in three slots, or, 3P3 = 6 . Binomial Distribution

Permutations & Combinations The general formula for a permutation is: When n = r, then nPn = n! n! is read as n factorial. In general, n! = n⋅(n-1)⋅(n-2)⋅ … ⋅(2)⋅(1) = n⋅(n-1)! So, 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3,628,800 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720 Note that 0! ≡ 1. Binomial Distribution

Permutations & Combinations Example: How many ways can you assign five workers to five different tasks? Answer: 5P5= 5! = 120 Example: How many ways can you arrange 10 different books in your bookcase which has room for exactly 5 books? Answer: 10P5= 10 x 9 x 8 x 7 x 6 = 30,240 Example: How many ways can 8 cars line up single file in front of a toll booth? Answer: 8P8 = 8! = 40,320 Example: How many ways can you arrange 12 guests around a table that has 12 chairs? Answer: 12P12 = 12! = 479,000,600 Follow up question: How many different potential family feuds can break out? Binomial Distribution

Permutations & Combinations With permutations, the arrangement of the items is important. Each unique sequence is another permutation. Thus, ABC is different from BCA and both are different from CBA. With combinations, however, ABC, BCA, and CBA are all the same. They are all part of the same combination. Example: How many different groups of 3 can be selected from 7 people? Say these people are named A, B, C, D, E, F, G. Note that once you select, say, B, D, and E, the six different arrangements you can make from them are irrelevant. Binomial Distribution

Permutations & Combinations The formula for combinations is: Note that: You can use the nCr key on your calculator to solve any combination problem. To solve the above problem: How many different groups of 3 can be selected from 7 people? Answer: 7C3 = 7! / 3! 4! = 35 Binomial Distribution

Permutations & Combinations Example: How many different hands can one draw from a deck of 52 cards in a game of 7-card rummy? Answer: 52C7 = 52! / 7! 45! = 133,784,560 Example: How many samples of size n = 6 can be drawn from a population of size N = 50? Answer: 50C6 = 50! / 6! 44! = 15,890,700 Binomial Distribution

Binomial Distribution The Binomial Distribution is used when the sampling process works this way: There are only two possible outcomes for each trial, object, or observation. These two outcomes are mutually exclusive. These outcomes are called success and failure. For example: heads/tails; pass/fail; defective/non-defective; dead/alive; hit/miss The outcomes are independent events.  The probability of success (or failure) is constant from trial to trial. For example, the probability of getting a head on a coin toss is the same at every toss of the coin. [This sampling process is called a Bernoulli process.] Binomial Distribution

Binomial Distribution The binomial distribution can be used to compute the probability of getting a particular number of successes in a specified number of trials (or observations). For: p ≡ the probability of success (1-p) ≡ the probability of failure x ≡ number of successes (n – x) ≡ number of failures n ≡ number of observations (or trials) The probability of getting x successes out of n trials (or observations) is: P(x) = nCx px (1-p)n-x Binomial Distribution

Binomial Distribution Again, the probability of getting x successes out of n trials (or observations) is: P(x) = nCx px (1-p)n-x The mean (expected value) of a binomial distribution is: μ = np The variance of a binomial distribution is: σ2 = np(1–p) Binomial Distribution

Example: Air conditioners An air conditioner made by a certain company is made of 20 distinct parts. Each part has a .004 probability of being defective. What is the probability that a randomly selected air conditioner will not work perfectly? [Note that it will not work perfectly if even one of these parts is defective.] Binomial Distribution

Example: Air conditioners Answer: First, solve the probability that the air conditioner will work perfectly, i.e., all 20 parts will be good. P(0 defectives) = P (X=20; n=20) = 20C20 (.004)0 (.996)20 = .923 The probability that the air conditioner will not work perfectly (i.e., that it has 1 or more defective parts) is: 1 - .923 = .077. There is a 7.7% chance that a randomly selected a/c made by this company will not work perfectly. What should be done to improve quality at this company? One solution is to reduce the number of parts. Also, improve quality of each part. .004 defective rate is not very good. [You will learn about 6 sigma quality in other courses.] Binomial Distribution

Example: Machine parts A machine produces parts that are very difficult to make. It turns out that 1 out of 20 are defective and must be thrown out. What is the probability that a sample of 10 parts will contain 0 defectives? Note: If 1 out of 20 parts are defective, this means that p = .05. (It is strange to call a defective part a “success”). Solution: P (x=0; n=10) = 10C0 (.05)0 (.95)10 =.5987 Binomial Distribution

Example: Coin toss If you toss a coin 12 times, what is the probability that you will get 6 heads? Solution: P (x=6; n=12) = 12C6 (.50)6 (.50)6 =.2256 The mean of the binomial, μ = np. On average, when you toss a coin 12 times, you expect to get 6 heads. The most likely outcome in 12 tosses is 6 heads. In 100 tosses, 50 heads is the most likely outcome, but the probability of getting 50 heads in 100 tosses is only .0796. Check for yourself: P (x=50; n=100) = 100C50 (.50)50 (.50)50 Binomial Distribution

Example: Females @ CUNY 60% of the students at CUNY are female. What is the probability that a randomly selected group of 25, there will be exactly 15 females? Solution: P (x=15; n=25) = 25C15 (.60)15 (.40)10 = .1612. Binomial Distribution