AQR Reading: Binomial Probability Reading #1: The Binomial Distribution A. It would be very tedious if, every time we had a slightly different problem, we had to determine the probability distributions from scratch. Luckily, there are enough similarities between certain types, or families, of experiments, to make it possible to develop formulas representing their general characteristics. For example, many experiments share the common element that their outcomes can be classified into one of two events, e.g. a coin can come up heads or tails; a child can be male or female; a person can die or not die; a person can be employed or unemployed. These outcomes are often labeled as success or failure. Note that there is no connotation of goodness here - for example, when looking at births, the statistician might label the birth of a boy as a success and the birth of a girl as a failure, but the parents wouldn t necessarily see things that way. The usual notation is p = probability of success, q = probability of failure = 1 - p. Note that p + q = 1. In statistical terms, A Bernoulli trial is each repetition of an experiment involving only 2 outcomes. We are often interested in the result of independent, repeated bernoulli trials, i.e. the number of successes in repeated trials. 1. independent - the result of one trial does not affect the result of another trial. 2. repeated - conditions are the same for each trial, i.e. p and q remain constant across trials. Hayes refers to this as a stationary process. If p and q can change from trial to trial, the process is nonstationary. The term identically distributed is also often used. B. A binomial distribution gives us the probabilities associated with independent, repeated Bernoulli trials. In a binomial distribution the probabilities of interest are those of receiving a certain number of successes, r, in n independent trials each having only two possible outcomes and the same probability, p, of success. So, for example, using a binomial distribution, we can determine the probability of getting 4 heads in 10 coin tosses. How does the binomial distribution do this? Basically, a two part process is involved. First, we have to determine the probability of one possible way the event can occur, and then determine the number of different ways the event can occur. That is, P(Event) = (Number of ways event can occur) * P(One occurrence). Suppose, for example, we want to find the probability of getting 4 heads in 10 tosses. In this case, we ll call getting a heads a success. Also, in this case, n = 10, the number of successes is r = 4, and the number of failures (tails) is n r = 10 4 = 6. One way this can occur is if the first 4 tosses are heads and the last 6 are tails, i.e.
S S S S F F F F F F The likelihood of this occurring is P(S) * P(S) * P(S) * P(S) * P(F) * P(F) * P(F) * P(F) * P(F) * P(F) More generally, if p = probability of success and q = 1 p = probability of failure, the probability of a specific sequence of outcomes where there are r successes and n-r failures is n r p r q So, in this particular case, p = q =.5, r = 4, n-r = 6, so the probability of 4 straight heads followed by 6 straight tails is.5 4.5 6 = 0.0009765625 (or 1 out of 1024). Of course, this is just one of many ways that you can get 4 heads; further, because the repeated trials are all independent and identically distributed, each way of getting 4 heads is equally likely, e.g. the sequence S S S S F F F F F F is just as likely as the sequence S F S F F S F F S F. So, we also need to know how many different combinations produce 4 heads. Well, we could just write them all out but life will be much simpler if we take advantage of two counting rules: 1. The number of different ways that N distinct things may be arranged in order is N! = (1)(2)(3)...(N-1)(N), (where 0! = 1). An arrangement in order is called a permutation, so that the total number of permutations of N objects is N!. The symbol N! is called N factorial. EXAMPLE. Rank candidates A, B, and C in order. The possible permutations are: ABC ACB BAC BCA CBA CAB. Hence, there are 6 possible orderings. Note that 3! = (1)(2)(3) = 6. 2. The total number of ways of selecting r distinct combinations of N objects, irrespective of order, is N! N N = = r!(n - r)! r N r We refer to this as N choose r. Sometimes the number of combinations is known as a binomial coefficient, and sometimes the notation N C r is used. So, in the present example,
Reading/Examples #2: The Binomial Distribution Formula: Binomial Distribution Overview The binomial distribution is a type of distribution in statistics that has two possible outcomes (the prefix bi means two, or twice). For example, a coin toss has only two possible outcomes: heads or tails and taking a test could have two possible outcomes: pass or fail (for more information, see What is a Binomial distribution?). A Binomial Distribution shows either (S)uccess or (F)ailure. Binomial distributions must also meet the following three criteria: 1. The number of observations or trials is fixed. In other words, you can only figure out the probability of something happening if you do it a certain number of times. This is common sense if you toss a coin once, your probability of getting a tails is 50%. IF you toss a coin a 20 times, your probability of getting a tails is very, very close to 100%. 2. Each observation or trial is independent. In other words, none of your trials have an effect on the probability of the next trial. 3. The probability of success (tails, heads, fail or pass) is exactly the same from one trial to another. Once you know that your distribution is binomial, you can apply the binomial distribution formula to calculate the probability. The Binomial Distribution Formula The binomial distribution formula is: b(x; n, P) = nc x * P x * (1 P) n x Where: b = binomial probability x = total number of successes (pass or fail, heads or tails etc.) P = probability of a success on an individual trial n = number of trials Note: The binomial distribution formula can also be written in a slightly different way, because nc x = n!/x!(n-x)! (this binomial distribution formula uses factorials. q in this formula is just the probability of failure (subtract your probability of success from 1).
Sample Problem Using the First Binomial Distribution Formula Q. A coin is tossed 10 times. What is the probability of getting exactly 6 heads? I m going to use this formula: b(x; n, P) nc x * P x * (1 P) n x The number of trials (n) is 10 The odds of success ( tossing a heads ) is 0.5 (So 1-p = 0.5) x = 6 P(x=6) = 10C 6 * 0.5^6 * 0.5^4 = 210 * 0.015625 * 0.0625 = 0.205078125 How to Work a Binomial Distribution Formula: Example #2 The binomial distribution formula can calculate the probability of success for binomial distributions. Often you ll be told to plug in the numbers to the formula and calculate. This is easy to say, but not so easy to do unless you are very careful with order of operations, you won t get the right answer. If you have a Ti-83 or Ti-89, the calculator can do much of the work for you. If not, here s how to break down the problem into simple steps so you get the answer right every time. Step 1:: Read the question carefully. Sample question: 80% of people who purchase pet insurance are women. If 9 pet insurance owners are randomly selected, find the probability that exactly 6 are women. Step 2:: Identify n and X from the problem. Using our sample question, n (the number of randomly selected items) is 9, and X (the number you are asked to find the probability for) is 6. Step 3: Work the first part of the formula. The first part of the formula is n! / (n X)! X! Substitute your variables: 9! / ((9 6)! 6!) Which equals 84. Set this number aside for a moment. Step 4: Find p and q. p is the probability of success and q is the probability of failure. We are given p = 80%, or.8. So the probability of failure is 1.8 =.2 (20%). Step 5: Work the second part of the formula. p X =.8 6 =.262144 Set this number aside for a moment. Step 6: Work the third part of the formula. (n X) q =.2 (9-6) =.2 3 =.008 Step 7: Multiply your answer from step 3, 5, and 6 together. 84.262144.008 = 0.176.
Reading #3: If the distribution of a random variable, x, fulfills the following requirements, then it is referred to as a "binomial distribution." We will learn how to compute binomial probabilities and characteristics of the binomial distribution. There are four requirements for a binomial experiment: 1. Each trial must have exactly TWO categories for outcomes (success and failure). 2. Experiment must have a fixed number of trials. (n) 3. Each trial must be independent of the others. 4. The probabilities for each trial must remain constant. p - probability of a success q = probability of a failure Success and failure are complementary events so p * Q = 1. The binomial formula states... P(x) =,C,'p"'qn-*! n is the number of trials (must be fixed). x is the number of successes out of n trials (remember that x may be any whole number between 0 and n, inclusive: x = 0, 1, 2... n). p is the probability of success on any given trial. q is the probability of failure on any given trial (q = I - p). P(x) is the probability of getting exactly x successes out of n trials., C" should be done on your calculator if possible. If not... nc, n! = (n - x)l xl Problems: Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. 1. Which of the following is not a property of a Binomial Experiment? a. All trials are identical. b. Each trial has only two possible outcomes. c. The probability of success may change from trial to trial. d. The purpose of the experiment is to determine the number of successes that occurs during the n trials. 2. In the expression, which value represents the number of trials? a. 2 c. 5 b. 3 d. 8 3. In the expression, which value represents the probability of failure? a. 0.6 c. (0.4) 2 b. 0.4 d. (0.6) 5 4. In the expression, which value represents the number of successes? a. 3 c. 5 b. 10 d. 7 5. Which expression describes the probability of k 3s being rolled on 20 successive rolls of a six-sided die?
b. 10 d. 7 5. Which expression describes the probability of k 3s being rolled on 20 successive rolls of a six-sided die? a. c. b. d. 6. The probability of a computer memory chip being defective is 0.02. Which of the following statements is true? a. In a shipment of 100 chips, two will be defective. b. The expected number of defective chips in a shipment of 500 is ten. c. In a shipment of 1000 chips, it is certain that at least one will be defective. d. All statements above are false. 7. A young couple plans to have a family with four children. Assuming that the behaviour of their first child does not cause them to alter their plans, what is the expected number of girls for their family? a. 2.5 c. 2 b. 2.25 d. 1.5 #8-12: Please show work below or on scratch paper for additional space 8. A hockey goaltender has a save percentage of 0.920. This means that the probability of any single shot taken on the goaltender being a goal is 0.08. What would be the expected number of goals scored on this goaltender in a game where she faced 35 shots? 9. A manufacturer of halogen bulbs knows that 3% of the production of their 100 W bulbs will be defective. What is the probability that exactly 5 bulbs in a carton of 144 bulbs will be defective? 10. A fair die has four faces numbered one to four. What is the probability of rolling a two exactly three times in ten rolls of the die? 11. A packet of carrot seeds has a germination rate of 92%. In other words, the probability of any seed sprouting is 0.92. How many seedlings would you expect in a row of 50 seeds? 12. A packet of vegetable seeds has a germination rate of 96%. What is the probability that exactly 10 of 12 seeds planted will sprout?