6 3 Binomial and Geometric Random Variables Objectives

6. 3 Binomial and Geometric Random Variables Objectives SWBAT: • DETERMINE whether the conditions for using a binomial random variable are met. • COMPUTE and INTERPRET probabilities involving binomial distributions. • CALCULATE the mean and standard deviation of a binomial random variable. INTERPRET these values in context. • FIND probabilities involving geometric random variables.

What are the conditions for a binomial setting? A binomial setting arises when we perform several independent trials of the same chance process and record the number of times that a particular outcome occurs. The four conditions for a binomial setting are: • Binary? The possible outcomes of each trial can be classified as “success” or “failure. ” • Independent? Trials must be independent; that is, knowing the result of one trial must not tell us anything about the result of any other trial. • Number? The number of trials n of the chance process must be fixed in advance. • Success? There is the same probability p of success on each trial.

What is a binomial random variable? What are the possible values of a binomial random variable? • When the same chance process is repeated several times, we are often interested in whether a particular outcome does or doesn’t happen on each repetition. Some random variables count the number of times the outcome of interest occurs in a fixed number of repetitions. They are called binomial random variables. • Some examples of binomial random variables: • Toss a coin 5 times. Count the number of heads. • Spin a roulette wheel 8 times. Record how many times the ball lands in a red slot. • Take a random sample of 100 cats. Count the number of females. • Binomial variables are discrete!

What are the parameters of a binomial distribution? • Consider tossing a coin n times. Each toss gives either heads or tails. Knowing the outcome of one toss does not change the probability of an outcome on any other toss. • If we define heads as a success, then p is the probability of a head and is 0. 5 on any toss. • The number of heads in n tosses is a binomial random variable X. The probability distribution of X is called a binomial distribution. The count X of successes in a binomial setting is a binomial random variable. The probability distribution of X is a binomial distribution with parameters n and p, where n is the number of trials of the chance process and p is the probability of a success on any one trial. The possible values of X are the whole numbers from 0 to n.

What is the most common mistake students make on binomial distribution questions? • On many questions involving binomial settings, students do not recognize that using the binomial distribution is appropriate. In fact, free response questions about the binomial distribution are often among the lowest scoring questions on the exam. • You need to spend time making sure you are able to identify a binomial distribution. If you aren’t sure how to answer a probability question, check if are working with a binomial setting.

Dice, Cars, and Hoops Determine whether the random variables below have a binomial distribution. Justify your answer. a) Roll a fair die 10 times and let X = the number of sixes. Binary? Yes, success = six, failure = not a six. Independent? Yes, knowing the outcomes of past rolls doesn’t provide additional information about the outcomes of future rolls. Number? Yes, there are 10 trials. Success? Yes, the probability of success is always 1/6. This is a binomial setting. X is binomial with n = 10 and p = 1/6.

b) Shoot a basketball 20 times from various distances on the court. Let Y = number of shots made. Binary? Yes, success = make the shot, failure = miss the shot. Independent? Yes, evidence suggests it is reasonable to assume that making a shot doesn’t change the probability of making the next shot. Number? Yes, there are 20 trials. Success? No, the probability of success changes because the shots are taken from various distances. This is not a binomial setting. Y is not binomial. c) Observe the next 100 cars that go by and let C = color. Binary? No. There are more than two possible colors. Also, C is not even a random variable since the outcomes aren’t numerical.

Rolling Sixes: In many games involving dice, rolling a 6 is desirable. The probability of rolling a six when rolling a fair die is 1/6. If X = the number of sixes in 4 rolls of a fair die, then X is binomial with n = 4 and p =1/6. What is P(X = 0)? That is, what is the probability that all 4 rolls are not sixes. What is P(X = 1)?

In general, how can we calculate binomial probabilities? Is the formula on the formula sheet? Binomial Probability There is a variant of this on the formula sheet. If X has the binomial distribution with n trials and probability p of success on each trial, the possible values of X are 0, 1, 2, …, n. If k is any one of these values, Number of arrangements of k successes Probability of n-k failures This is the binomial coefficient: the number of ways that exactly k successes can occur in a set of n trials.

• Let’s go back to the die example and look at how we can use this formula when X = 3. Reminder, this probability was 0. 015.

The binompdf function calculates these probabilities for us. To locate, go to 2 nd, DISTR, A: binompdf(n, p, k) computes P(X = k) n is your number of trials, p is the probability of success, and k is the x value In the previous example, we saw that the probability of rolling exactly three 6’s was 0. 015. On the calculator, we would enter: binompdf(trials: 4, p: 1/6, x value: 3) We would get 0. 015.

Let’s try another example. At a certain intersection, the light for eastbound traffic is red for 15 seconds, yellow for 5 seconds, and green for 30 seconds. Find the probability that out of the next eight eastbound cars that arrive randomly at the light, exactly three will be stopped by a red light. Let X = the number of stops at a red light n = 8 and p = 15/50 (note: success is a red light, failure is any other color) Technology:

Try this problem: Each child of a particular set of parents has probability 0. 25 of having type O blood. Suppose the parents have 5 children. Find the probability that exactly 3 of the children have type O blood. Let X = the number of children with type O blood. n = 5 and p =0. 25

• It seems like there would be a more convenient way to do this…

When working with a binomial distribution, we won’t always find probabilities for an exact X value. Sometimes we will find probabilities less than an X value and sometimes we will find probabilities greater than an X value. These require a different formula and different command on the calculator. • The command used in this situation would be option B in DISTR for binomcdf. It is okay to use binompdf and binomcdf commands on the AP exam! However, you need to keep a few things in mind: Step 1: State the distribution and the values of interest. Specify a binomial distribution with the number of trials n, success probability p, and the values of the variable clearly identified. Step 2: Perform calculations—show your work! Do one of the following: (i) Use the binomial probability formula to find the desired probability; or (ii) Use binompdf or binomcdf command label each of the inputs. Step 3: Answer the question.

Let’s revisit: Each child of a particular set of parents has probability 0. 25 of having type O blood. Suppose the parents have 5 children. Find the probability that the parents have less than 2 children with type O blood. When we did this before, we got 0. 6328. Each child of a particular set of parents has probability 0. 25 of having type O blood. Suppose the parents have 5 children. Find the probability that the parents have more than 3 children with type O blood. We got 0. 0156 before.

Roulette: In Roulette, 18 of the 38 spaces on the wheel are black. Suppose you observe the next 10 spins of a roulette wheel. a) What is the probability that exactly 4 of the spins land on black? Let X = the number of times the ball lands in a black slot There are 10 independent trials of the chance process, each with success probability 18/38. So X has a binomial distribution with n =10 and p = 18/38. To find P(X = 4): binompdf(trials: 10, p: 18/38, x value: 4) = 0. 225 There is a 0. 225 probability that exactly 4 of the 10 spins land on black.

Roulette: In Roulette, 18 of the 38 spaces on the wheel are black. Suppose you observe the next 10 spins of a roulette wheel. b) What is the probability that at least 8 of the spins land on black? Let Y = the number of times the ball lands in a black slot There are 10 independent trials of the chance process, each with success probability 18/38. So Y has a binomial distribution with n =10 and p = 18/38. 1 - binomcdf(trials: 10, p: 18/38, x value: 7) = 1 – 0. 9615 = 0. 0385 There is a 0. 0385 probability that at least 8 of the 10 spins land on black.

How can you calculate the mean and SD of a binomial distribution? Are these on the formula sheet? Mean and Standard Deviation of a Binomial Random Variable If a count X has the binomial distribution with number of trials n and probability of success p, the mean and standard deviation of X are These are on the formula sheet.

Roulette: Let X = the number of the next 10 spins of a roulette wheel that land on black. a) Calculate and interpret the mean and standard deviation of X. n = 10 and p = 18/38 If many individuals span a roulette wheel 10 times, the average number of spins to land on black would be 4. 7368. If many individuals span a roulette wheel 10 times, the number of spins to land on black would typically vary by about 1. 5789 spins from the mean (4. 7368).

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When is it okay to use a binomial distribution when sampling without replacement? Why is this an issue? • The binomial distributions are important in statistics when we wish to make inferences about the proportion p of successes in a population. • Almost all real-world sampling, such as taking an SRS from a population of interest, is done without replacement. However, sampling without replacement leads to a violation of the independence condition. • When the population is much larger than the sample, a count of successes in an SRS of size n has approximately the binomial distribution with n equal to the sample size and p equal to the proportion of successes in the population. 10% Condition When taking an SRS of size n from a population of size N, we can use a binomial distribution to model the count of successes in the sample as long as

• What this means is that the binomial distribution is a good approximation as long as we don’t sample more than 10% of our population • Do not mistake this to mean that we want a small sample. This is almost never the case. It just means that if we have a sample that is larger than 10% of the population, we shouldn’t use the binomial distribution. Almost everyone has one—a drawer that holds miscellaneous batteries of all sizes. Suppose that your drawer contains 8 AAA batteries but only 6 of them are good. You need to choose 4 for your graphing calculator. If you randomly select 4 batteries, what is the probability that all 4 of them will work? On the calculator this is binompdf (trials: 4, p: 6/8, x value: 4) = 0. 3164 The problem is that we are taking a sample of 4 batteries from a population of 8 batteries, so our sample size is 50% of our population, which violates the 10% condition. Therefore, a binomial distribution is not an appropriate distribution to use. The real probability here is:

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• In a binomial setting, the number of trials n is fixed in advance, and the binomial random variable X counts the number of successes. The possible values of X are 0, 1, 2, …, n. In other situations, the goal is to repeat a chance process until a success occurs. • Roll a pair of dice until you get doubles • In basketball, attempt a three-point shot until you make one. • Keep placing a $1 bet on the number 7 in roulette until you win. • These are all examples of a geometric setting. • Although the number of trials isn’t fixed in advance (we don’t know how long it will take until we achieve a success), the trials are independent and the probability of success remains constant.

What are the conditions for a geometric setting? • A geometric setting arises when we perform independent trials of the same chance process and record the number of trials it takes to get one success. On each trial, the probability p of success must be the same. What is a geometric random variable? What are the possible values of a geometric random variable? What are the parameters of a geometric distribution? • The number of trials Y that it takes to get a success in a geometric setting is a geometric random variable. The probability distribution of Y is a geometric distribution with parameter p, the probability of a success on any trial. The possible values of Y are 1, 2, 3, . . Like binomial random variables, it is important to be able to distinguish situations in which the geometric distribution does and doesn’t apply!

Monopoly: In the board game Monopoly, one way to get out of jail is to roll doubles. Suppose that a player has to stay in jail until he or she rolls doubles. The probability of rolling doubles is 1/6. a) Explain why this is a geometric setting. The random variable of interest in this example is Y = number of attempts it takes to roll doubles one time. Each attempt is one trial of the chance process. Knowing the outcome of previous rolls does not tell us anything about future rolls (trials are independent). On each roll, the probability of success is 1/6. This is a geometric setting. Because Y counts the number of attempts it takes to get doubles, it is a geometric random variable with parameter p = 1/6. b) Define the geometric random variable and state its distribution. Y = number of attempts it takes to roll doubles one time Geometric distribution with parameter p = 1/6.

c) Find the probability it takes exactly three rolls to get out of jail. Exactly four rolls. 100 rolls.

In general, how can you calculate geometric probabilities? Is this formula on the formula sheet? Geometric Probability Formula If Y has the geometric distribution with probability p of success on each trial, the possible values of Y are 1, 2, 3, …. If k is any one of these values, On calculator: This is NOT on the formula sheet

On average, how many rolls should it take to escape jail in Monopoly? What do you think? The probability of rolling doubles is 1/6. So on average, how many rolls should it take before you roll doubles? 6! In general, how do you calculate the mean of a geometric distribution? Is the formula on the formula sheet? Mean (Expected Value) Of A Geometric Random Variable If Y is a geometric random variable with probability p of success on each trial, then its mean (expected value) is E(Y) = µY = 1/p.

What is the probability it takes longer than average to escape jail? What does this probability tell you about the shape of the distribution? We want to find P(Y>6). However, there an infinite number of possible values for Y greater than 6. We need to use our complement rule. The shape of the distribution is skewed right, as all geometric distributions are. The probabilities in the beginning are very high, and then spread out as you move to the right.