A.1 Geometric Distribution
Learning Objectives
By the end of this section, you will be able to:
- list the characteristics that make an experiment geometric, and tell a geometric setting apart from a binomial one;
- explain what it means for the geometric distribution to be memoryless, and apply that property to a run of past failures;
- compute \(P(X = x)\) for a geometric random variable, and use the tail formulas to answer "more than" and "less than" questions;
- calculate the mean and standard deviation of a geometric distribution, and say in plain words what each one measures;
- distinguish the two forms of the distribution — counting trials until the first success versus counting failures before it — and pick the right formulas for each.
A.1.1 What Makes an Experiment Geometric
A geometric experiment is a statistical experiment with all four of the following properties.
- A trial is repeated until a success occurs. Think of it as a string of Bernoulli trials that are all failures except the last one, which is a success. You keep repeating what you are doing until the first success, and then you stop. In theory the number of trials could go on forever.
- The repeated trials are independent of each other.
- The probability \(p\) of a success and the probability \(q\) of a failure are the same on every trial, with \(p + q = 1\) and \(q = 1 - p\).
- The random variable \(X\) counts the trial on which the first success occurs. That is, \(X\) is the number of independent trials up to and including the first success.
Definition A.1.1 — The run stops the moment the first success lands, and X counts every trial up to and including it.
A distribution is memoryless when the probability of waiting \(n\) more trials for a success does not depend on how many failures have already occurred. Formally, for a geometric random variable \(X\),
$$P(X = n + k \mid X \geq k + 1) = P(X = n)$$where \(k\) is the number of previous failures.
Definition A.1.2 — Ten misses of history and no history at all give the batter the exact same next three at bats.
Read those four properties as a checklist: if a problem satisfies all four, the formulas in this section apply. Characteristic 1 is the one that separates this from Section 4.3. A binomial experiment fixes the number of trials in advance and lets the number of successes vary; a geometric experiment fixes the number of successes at one and lets the number of trials vary. You throw a dart at a bullseye until you hit it. The first time you hit it is a success, so you stop throwing. It might take six tries: failure, failure, failure, failure, failure, success, STOP.
Characteristic 3 is worth testing on a concrete case. The probability of rolling a three with one fair die is \(p = \frac{1}{6}\), and that is true no matter how many times you have already rolled. Suppose you want the probability of getting your first three on the fifth roll. Rolls one through four must all fail, each with probability \(\frac{5}{6}\), and roll five must succeed with probability \(\frac{1}{6}\):
$$P(X = 5) = \left(\frac{5}{6}\right)^{4}\left(\frac{1}{6}\right) \approx 0.0804$$Beyond the four defining characteristics, the geometric distribution carries three more attributes worth naming.
The random variable is discrete, and it can be defined in two ways. For the die, we asked "what is the probability the first success is on the fifth throw?" We could instead have asked "what is the probability it takes four failures before a success?" Those are the same experiment counted two different ways, and each way changes the form of the probability function slightly, along with the mean and the standard deviation. We handle both forms in the last part of this section.
The probability of success is assumed constant, which is a modeling choice. For flipping a coin that is obvious. For experiments that require skill — hitting a baseball, throwing a dart — you might reasonably object that the thrower is learning as the experiment runs, so the odds should improve. The geometric distribution cannot capture learning. It uses the historical probability of success and holds it fixed.
The geometric distribution is memoryless. Very few probability density functions have this property, and among discrete random variables the geometric distribution is the only one.
A coin that has landed tails nine times in a row is not "due" for heads. The geometric distribution builds that fact in: past failures carry no information about the next trial, so the count restarts every time you step up to try again.
Here is what that looks like in practice. Historically, Major League Baseball player Jones reaches at least first base with probability 0.20. Jones has not had a hit in his last 10 at bats. What is the probability Jones gets a hit on his third at bat from now? The answer ignores all 10 previous failures. Everything before the current moment is irrelevant — that is what "memoryless" means.
Jones's probability of a hit begins fresh every time he comes to the plate. This leads to a result that surprises people: if you are pulling parts off a manufacturing line to test for defects, the geometric distribution starts with a clean slate each time you begin testing, with no consideration of how the previous batch went. You will meet this property again when we get to the exponential probability density function.
You throw darts at a board until you hit the center area. Your probability of hitting the center area is \(p = 0.17\). You want to find the probability that it takes eight throws until you hit the center. What values does \(X\) take on?
Solution
Step 1 — Name the random variable. Let \(X\) be the number of throws it takes until you hit the center area, counting the successful throw.
Step 2 — List the values. The first throw could hit, so \(X\) can be 1. It could take two throws, or three, and there is no upper limit on how long a run of misses could last.
Answer: \(X\) takes on the values 1, 2, 3, ... — every positive whole number, with no fixed upper bound.
Marcus and his husband play a game of chance that a player can either win or lose (there are no other possibilities), and Marcus keeps playing until he loses. His probability of losing any one game is \(p = 0.57\). What is the probability that it takes five games until he loses?
Let \(X\) be the number of games Marcus plays until he loses, including the losing game. Then \(X\) takes on the values 1, 2, 3, ... and could in principle go on indefinitely. The probability question is \(P(X = 5)\).
Solution
Step 1 — Identify success and failure. The event we are waiting for is Marcus losing the game, so a "success" here is a loss, with \(p = 0.57\). A "failure" is a win, with \(q = 1 - 0.57 = 0.43\). The labels feel backwards, but the arithmetic only cares which outcome ends the experiment.
Step 2 — Describe the sequence. For it to take exactly five games, his first four games must be wins and the fifth must be a loss.
Step 3 — Multiply, using independence.
$$P(X = 5) = (0.43)^{4}(0.57) \approx 0.0195$$Answer: \(P(X = 5) \approx 0.0195\), so a bit under a 2% chance.
A.1.2 Setting Up and Solving Geometric Problems
Most geometric questions are not "what is the probability it takes exactly \(x\) tries." They are "on average, how long will this take?" or "what is the chance it takes at least three tries?" Both have short formulas, and the second one needs a trick.
The mean of a geometric distribution is \(\mu = \frac{1}{p}\), and the intuition is clean: if something works one time in five, expect to wait about five tries. The "at least" question is harder, because a geometric random variable runs to infinity — you cannot add up every term. The way around it is to add up the short side instead. Asking for the probability that it takes more than \(n\) trials is the same as asking for the probability that the first \(n\) trials all fail, and that is just \((1-p)^n\). Everything else is the complement of that.
With a binomial variable you could add the individual probabilities from 0 up to 3 and be done. A geometric variable has infinitely many values, so you can never finish the sum. Counting the failures instead turns an infinite sum into one multiplication.
The two formulas below package that idea. Read \(n\) as the cutoff the question names, and notice that the first one is nothing more than "the first \(n\) trials all failed."
What is the probability it takes MORE THAN \(n\) trials to get the first success?
$$P(X > n) = (1 - p)^{n}$$What is the probability it takes LESS THAN OR EQUAL TO \(n\) trials?
$$P(X \leq n) = 1 - (1 - p)^{n}$$Because \(X\) is a whole number, \(P(X \geq n) = P(X > n - 1) = (1-p)^{n-1}\).
Professor Avery Hollis feels that 15% of their students get below a C on the final exam. They decide to look at final exams (selected randomly and replaced in the pile after reading) until they find one that shows a grade below a C. We want to know the probability that Professor Hollis will have to examine at least ten exams until they find one with a grade below a C. What is the probability question stated mathematically?
Solution
Step 1 — Name the random variable. Let \(X\) be the number of final exams Professor Hollis must examine until they find one with a grade below a C.
Step 2 — Translate the words. "At least ten" means ten or more, which is the symbol \(\geq 10\).
Answer: The probability question is \(P(X \geq 10)\). If you wanted its value, it is \(P(X > 9) = (1 - 0.15)^{9} = (0.85)^{9} \approx 0.2316\).
Rosa Delgado, a safety engineer, feels that 35% of all industrial accidents in the plant are caused by failure of employees to follow instructions. She decides to look at the accident reports (selected randomly and replaced in the pile after reading) until she finds one that shows an accident caused by failure of employees to follow instructions.
a. On average, how many reports would Rosa expect to look at until she finds a report showing an accident caused by employee failure to follow instructions?
b. What is the probability that she will have to examine at least three reports until she finds one?
Let \(X\) be the number of accident reports Rosa must examine until she finds one caused by employee failure to follow instructions. \(X\) takes on the values 1, 2, 3, ... — unlike a binomial experiment with its fixed number of trials, a geometric experiment can rack up any number of failed trials before the success arrives.
Solution
Part a — "on average" means find the mean. The formula for the mean of this geometric distribution is:
$$\mu = E(X) = \frac{1}{p} = \frac{1}{0.35} \approx 2.86$$The mean does not have to be a whole number even though the random variable is discrete and must be a counting number. Rosa will always read a whole report; 2.86 is the long-run average over many repetitions of the whole procedure.
Part b — "at least" translates to \(\geq\). We want \(P(X \geq 3)\). Reading at least three reports means the first two both failed to show the cause we are looking for:
$$P(X \geq 3) = P(X > 2) = (1 - 0.35)^{2} = (0.65)^{2} = 0.4225$$Answer: a. About 2.86 reports on average. b. \(P(X \geq 3) = 0.4225\), so a little better than a 42% chance.
Check an “at least” answer in rāSHio
Part b asks for \(P(X \geq 3)\), and rāSHio will confirm it: choose Distributions → Geometric, set the success probability to 0.35, pick Right, and enter \(k = 2\) — one less than the three reports, because the tool counts the failures before the success rather than the reports read. It returns 0.4225, the number the tail formula gives. It will not hand you part a, though: the panel reports probabilities only, so the mean \(\frac{1}{p}\) stays yours to compute.
Mateo Herrera needs to find a store that carries a special printer ink. He knows that of the stores that carry printer ink, 10% of them carry the special ink. He randomly calls each store until one has the ink he needs. What are \(p\) and \(q\)?
Solution
Step 1 — Decide what counts as a success. The event he is waiting for is a store having the special ink, and 10% of stores do.
Step 2 — Read off both probabilities. \(p = 0.10\), and \(q = 1 - 0.10 = 0.90\).
Answer: \(p = 0.10\) and \(q = 0.90\).
Ben Carlisle is looking for a student at his college who lives within five miles of him. He knows that 55% of the 25,000 students do live within five miles of him. He randomly contacts students from the college until one says they live within five miles of him.
This is a geometric problem because he may have a number of failures before he gets the one success he wants. The probability of a success stays the same each time he asks a student, and there is no fixed number of trials.
a. Let \(X\) be the number of _______ he must ask until one says yes.
b. What values does \(X\) take on?
c. What are \(p\) and \(q\)?
d. The probability question is \(P(\underline{\phantom{XXXX}})\), and what is its value?
Solution
Part a — fill in the blank. Let \(X\) be the number of students he must ask until one says yes.
Part b — the possible values. \(X\) takes on 1, 2, 3, ..., up to the total number of students. The first person Ben asks might say yes, or he might work his way through a long run of noes first.
Part c — the two probabilities. A success is a student saying yes, so \(p = 0.55\), and \(q = 1 - 0.55 = 0.45\).
Part d — the question and its value. The probability question is \(P(X = 4)\). For it to take exactly four students, the first three must say no and the fourth must say yes:
$$P(X = 4) = (0.45)^{3}(0.55) \approx 0.0501$$Answer: a. students; b. 1, 2, 3, ..., 25,000; c. \(p = 0.55\) and \(q = 0.45\); d. \(P(X = 4) \approx 0.0501\).
A.1.3 Notation, the Two Cases, and the Common Ratio
Let \(X\) be the number of independent trials up to and including the first success, so \(X = 1, 2, 3, \ldots\). Then
$$P(X = x) = (1 - p)^{x-1}\,p$$with mean and standard deviation
$$\mu = \frac{1}{p} \qquad\text{and}\qquad \sigma = \sqrt{\left(\frac{1}{p}\right)\left(\frac{1}{p} - 1\right)}$$Definition A.1.3 — Case I counts every trial including the success, which is why the exponent is one less than x.
Let \(X\) be the number of failures that occur before the first success, so \(X = 0, 1, 2, 3, \ldots\). Then
$$P(X = x) = (1 - p)^{x}\,p$$with mean and standard deviation
$$\mu = \frac{1 - p}{p} \qquad\text{and}\qquad \sigma = \sqrt{\frac{1 - p}{p^{2}}}$$The common ratio \(r\) of a geometric distribution is the constant factor between consecutive probabilities. It is found by dividing any probability by the one before it:
$$r = \frac{P(X = x + 1)}{P(X = x)} = 1 - p$$Same experiment, two different scoreboards. Case I counts every attempt including the one that worked; Case II counts only the misses. The counts always differ by exactly one, and so do the means.
Definition A.1.5 — The step from each probability to the next is the same multiplier every time, and that multiplier is 1 - p.
This is the case we have used all section: \(X\) counts trials, and the trial that succeeds is included in the count. The exponent \(x - 1\) is doing the bookkeeping. If it takes \(x\) trials in total and the last one succeeds, then exactly \(x - 1\) of them failed, so the failure probability \((1-p)\) gets raised to that power and the success probability \(p\) is multiplied in once. The mean \(\frac{1}{p}\) tells you how many trials to expect, counting the successful one.
The shorthand for a geometric distribution is \(X \sim G(p)\), read as "\(X\) is a random variable with a geometric distribution." The single parameter is \(p\), the probability of a success on each trial. That one number determines everything else — the whole probability function, the mean, and the standard deviation.
A geometric random variable can also be defined a second way, and being precise about the difference matters: the two definitions produce different formulas, and mixing them is the most common mistake in this section. The second version counts the misses only, which shifts the whole picture down by one.
Here the successful trial is not counted, so the exponent on \((1-p)\) is exactly \(x\), the number of failures. The mean \(\frac{1-p}{p}\) tells you how many failures to expect before the success arrives, which is always one less than the Case I mean.
In either case, the sequence of probabilities forms a geometric sequence — each term is a fixed multiple of the one before it. That is where the distribution gets its name, and it is a property no other probability distribution has.
Figure A.1.1 — The geometric distribution for p = 0.02, whose bars decline by the same factor at every step.
Figure A.1.1 graphs the probabilities for a defect rate of \(p = 0.02\). The vertical axis is the probability of \(X\), and the horizontal axis is the number of the component being tested. Using the Case I formula: at \(x = 1\) the probability the very first component tested is defective is \(P(X = 1) = 0.0200\). At \(x = 2\), the probability the second component is the first defective one is \(P(X = 2) = (0.98)(0.02) = 0.0196\), and for the third, \(P(X = 3) = (0.98)^{2}(0.02) = 0.019208\). Rounded to three decimal places the first two are both 0.020, which is why they look identical on the graph.
Notice on Figure A.1.1 that the probabilities decline by the same step down with every change in \(x\). That step is the common ratio.
For the defect data, \(\frac{P(X = 5)}{P(X = 4)} = \frac{0.018447}{0.018824} = 0.98\), and \(r = 1 - 0.02 = 0.98\) confirms it. Multiplying any probability by the common ratio gives you the next one in the sequence without going back to the formula. The probability that the sixth component tested is the first defect is 0.018078, so
$$P(X = 7) = (0.018078)(0.98) = 0.017716$$which matches the value the Case I formula gives directly. Once you have the common ratio, any \(P(X = x_a)\) you want is a short walk away.
The number of components you would expect to test until you find the first defective one is the mean, \(\mu = \frac{1}{p} = \frac{1}{0.02} = 50\). Its variance and standard deviation follow from the Case I formulas:
$$\sigma^{2} = \left(\frac{1}{p}\right)\left(\frac{1}{p} - 1\right) = \left(\frac{1}{0.02}\right)\left(\frac{1}{0.02} - 1\right) = 2{,}450$$ $$\sigma = \sqrt{2{,}450} \approx 49.5$$A standard deviation of 49.5 sitting next to a mean of 50 is not a mistake. When successes are rare the waiting time is wildly variable — you might find a defect on the third component or on the two-hundredth — and the geometric distribution says so plainly.
The probability of a defective steel rod is 0.01. Steel rods are selected at random. Find the probability that the first defect occurs on the ninth steel rod.
Solution
Step 1 — Set up the Case I formula. A defect on the ninth rod means the first eight were fine, with \(p = 0.01\):
$$P(X = 9) = (1 - 0.01)^{8}(0.01) = (0.99)^{8}(0.01)$$Step 2 — Evaluate. \((0.99)^{8} \approx 0.9227\), so \(P(X = 9) \approx 0.0092\). On a TI-83/84 this is geometpdf(0.01, 9).
Answer: \(P(X = 9) \approx 0.0092\).
Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested. How many components do you expect to test until one is found to be defective?
Let \(X\) be the number of computer components tested until the first defect is found. \(X\) takes on the values 1, 2, 3, ... with \(p = 0.02\).
Solution
Step 1 — Use the Case I formula. For the seventh component to be the first defect, the first six must be good:
$$P(X = 7) = (1 - 0.02)^{6}(0.02) = (0.98)^{6}(0.02) \approx 0.0177$$Step 2 — Find the expected number of tests.
$$\mu = \frac{1}{p} = \frac{1}{0.02} = 50$$Answer: The probability that the seventh component is the first defect is 0.0177. You expect to test 50 components before finding a defective one.
Try it in rāSHio — but count the misses, not the tries
Open rāSHio and choose Distributions → Geometric, then set the success probability to \(p = 0.02\). Here is the catch, and it is exactly the Case I / Case II split this section warns about: rāSHio's \(k\) counts the FAILURES before the first success, so for “the seventh component is the first defect” you enter \(k = 6\), not 7, and Exact returns 0.0177. The cumulative question shifts the same way — \(P(X \leq 7)\) is Left with \(k = 6\). Whenever you bring a Case I question to the tool, subtract one from the trial number before you type it.
Figure A.1.2 — Reading a geometric probability off the rāSHio panel: Distributions → Geometric. Its \(k\) is a failure count, so it runs one below this section's trial number. The walkthrough uses its own demo numbers — the steps are the ones above.
The literacy rate for a nation measures the proportion of people age 15 and over who can read and write. Kao Xiong is researching literacy for a global development class, and they find that the literacy rate for women in Afghanistan is 12%. Let \(X\) be the number of Afghani women Kao asks until one says that she is literate.
a. What is the probability distribution of \(X\)?
b. What is the probability that they ask five women before one says she is literate?
c. What is the probability that they must ask ten women?
d. Find the (i) mean and (ii) standard deviation of \(X\).
Solution
Part a — name the distribution. A success is a woman saying she is literate, with \(p = 0.12\), so \(X \sim G(0.12)\).
Part b — the fifth woman is the first yes.
$$P(X = 5) = (1 - 0.12)^{4}(0.12) = (0.88)^{4}(0.12) \approx 0.0720$$Part c — the tenth woman is the first yes.
$$P(X = 10) = (0.88)^{9}(0.12) \approx 0.0380$$Part d — mean and standard deviation.
$$\mu = \frac{1}{p} = \frac{1}{0.12} \approx 8.33$$ $$\sigma = \sqrt{\frac{1 - 0.12}{(0.12)^{2}}} = \sqrt{\frac{0.88}{0.0144}} \approx 7.82$$Answer: a. \(X \sim G(0.12)\); b. 0.0720; c. 0.0380; d. mean \(\approx 8.33\) women, standard deviation \(\approx 7.82\) women.
Nia Adeyemi and her wife volunteer for a community health survey. The lifetime risk of developing cancer is about one in 67 (1.5%). Let \(X\) be the number of people Nia asks until one says they have cancer. Then \(X\) is a discrete random variable with a geometric distribution: \(X \sim G\!\left(\frac{1}{67}\right)\), or \(X \sim G(0.015)\).
a. What is the probability that she asks ten people before one says they have cancer?
b. What is the probability that she must ask 20 people?
c. Find the (i) mean and (ii) standard deviation of \(X\).
Solution
Part a — the tenth person is the first yes.
$$P(X = 10) = (1 - 0.015)^{9}(0.015) \approx 0.0131$$On a calculator this is geometpdf(0.015, 10).
Part b — the twentieth person is the first yes.
$$P(X = 20) = (1 - 0.015)^{19}(0.015) \approx 0.0113$$Part c — mean and standard deviation. Using the Case I formulas with \(p = 0.015\):
$$\mu = \frac{1}{p} = \frac{1}{0.015} \approx 66.67$$ $$\sigma = \sqrt{\frac{1 - p}{p^{2}}} = \sqrt{\frac{1 - 0.015}{(0.015)^{2}}} \approx 66.16$$Answer: a. 0.0131; b. 0.0113; c. mean \(\approx 66.67\) people, standard deviation \(\approx 66.16\) people.
Problem Set A.1
Problem 1. Erin Blackwell, looking to buy a used red Miata, will call dealerships until she finds one that carries the car. She estimates the probability that any independent dealership will have the car will be 28%. We are interested in the number of dealerships she must call.
a) In words, define the random variable \(X\).
b) List the values that \(X\) may take on.
c) Give the distribution of \(X\). \(X \sim\) _______(_______)
d) On average, how many dealerships would we expect her to have to call until she finds one that has the car?
e) Find the probability that she must call at most four dealerships.
f) Find the probability that she must call three or four dealerships.
Solution
Step 1 — Define the random variable: Erin keeps calling dealerships until one of them has the car, then stops, so the count is a geometric one. Let \(X\) be the number of dealerships Erin must call until she reaches one that carries the used red Miata, counting the successful call.
Step 2 — List the possible values: The very first dealership might have the car, or she might work through a long run of dealerships that do not. There is no fixed upper limit, so \(X\) takes on 1, 2, 3, ... .
Step 3 — Name the distribution: A success is a dealership carrying the car, with probability \(p = 0.28\).
$$X \sim G(0.28)$$Step 4 — Find the expected number of calls: The mean of a geometric distribution is the reciprocal of the success probability.
$$\mu = \frac{1}{p} = \frac{1}{0.28} \approx 3.57$$Step 5 — Find \(P(X \leq 4)\): "At most four" is the complement of "the first four calls all failed," so use the tail formula.
$$P(X \leq 4) = 1 - (1 - 0.28)^{4} = 1 - (0.72)^{4} = 1 - 0.2687 \approx 0.7313$$Step 6 — Find \(P(X = 3 \text{ or } X = 4)\): These are two separate values of \(X\), so compute each with the Case I formula and add.
$$P(X = 3) = (0.72)^{2}(0.28) = 0.1452$$ $$P(X = 4) = (0.72)^{3}(0.28) = 0.1045$$ $$P(X = 3 \text{ or } 4) = 0.1452 + 0.1045 \approx 0.2497$$Answer: a) \(X\) = the number of dealerships Erin calls until one carries the car, including that call; b) 1, 2, 3, ...; c) \(X \sim G(0.28)\); d) about 3.57 dealerships; e) \(P(X \leq 4) \approx 0.7313\); f) \(P(X = 3 \text{ or } 4) \approx 0.2497\).
Problem 2. Suppose that the probability that an adult in America will watch the Super Bowl is 40%. Each person is considered independent. We are interested in the number of adults in America we must survey until we find one who will watch the Super Bowl.
a) In words, define the random variable \(X\).
b) List the values that \(X\) may take on.
c) Give the distribution of \(X\). \(X \sim\) _______(_______)
d) How many adults in America do you expect to survey until you find one who will watch the Super Bowl?
e) Find the probability that you must ask seven people.
f) Find the probability that you must ask three or four people.
Solution
Step 1 — Define the random variable: We survey adults one at a time and stop as soon as someone says they will watch, so this is a geometric count. Let \(X\) be the number of adults in America we must survey until we find one who will watch the Super Bowl, including that person.
Step 2 — List the possible values: The first person surveyed could be a viewer, or it could take many tries, with no fixed ceiling. So \(X\) takes on 1, 2, 3, ... .
Step 3 — Name the distribution: A success is finding a Super Bowl viewer, with probability \(p = 0.40\).
$$X \sim G(0.40)$$Step 4 — Find the expected number surveyed:
$$\mu = \frac{1}{p} = \frac{1}{0.40} = 2.5$$On average it takes two and a half people, which makes sense when two out of five adults watch.
Step 5 — Find \(P(X = 7)\): For the seventh person to be the first viewer, the first six must all be non-viewers.
$$P(X = 7) = (1 - 0.40)^{6}(0.40) = (0.6)^{6}(0.4) = (0.046656)(0.4) \approx 0.0187$$Step 6 — Find \(P(X = 3 \text{ or } X = 4)\): Compute each value separately and add.
$$P(X = 3) = (0.6)^{2}(0.4) = 0.1440$$ $$P(X = 4) = (0.6)^{3}(0.4) = 0.0864$$ $$P(X = 3 \text{ or } 4) = 0.1440 + 0.0864 = 0.2304$$Answer: a) \(X\) = the number of adults surveyed until one says they will watch the Super Bowl; b) 1, 2, 3, ...; c) \(X \sim G(0.40)\); d) 2.5 adults; e) \(P(X = 7) \approx 0.0187\); f) \(P(X = 3 \text{ or } 4) = 0.2304\).
Problem 3. It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose we are interested in the number of California residents we must survey until we find a resident who does not have adequate earthquake supplies.
a) In words, define the random variable \(X\).
b) List the values that \(X\) may take on.
c) Give the distribution of \(X\). \(X \sim\) _______(_______)
d) What is the probability that we must survey just one or two residents until we find a California resident who does not have adequate earthquake supplies?
e) What is the probability that we must survey at least three California residents until we find a California resident who does not have adequate earthquake supplies?
f) How many California residents do you expect to need to survey until you find a California resident who does not have adequate earthquake supplies?
g) How many California residents do you expect to need to survey until you find a California resident who does have adequate earthquake supplies?
Solution
Step 1 — Decide which outcome is the success: The problem asks how many residents we survey until we find one who does not have adequate earthquake supplies. Since 30% do have adequate supplies, the event we are waiting for has probability
$$p = 1 - 0.30 = 0.70$$Getting this backwards is the main trap in this problem — the success is always whatever event stops the survey.
Step 2 — Define the random variable: Let \(X\) be the number of California residents we must survey until we find one who does not have adequate earthquake supplies, including that resident.
Step 3 — List the possible values: \(X\) takes on 1, 2, 3, ... .
Step 4 — Name the distribution:
$$X \sim G(0.70)$$Step 5 — Find \(P(X = 1 \text{ or } X = 2)\): Either the first resident we survey is unprepared, or the first is prepared and the second is not.
$$P(X = 1) = 0.70 \qquad P(X = 2) = (0.30)(0.70) = 0.21$$ $$P(X = 1 \text{ or } 2) = 0.70 + 0.21 = 0.91$$Step 6 — Find \(P(X \geq 3)\): Surveying at least three residents means the first two were both prepared. Use the tail formula, or simply subtract from 1.
$$P(X \geq 3) = (1 - 0.70)^{2} = (0.30)^{2} = 0.09$$That agrees with \(1 - 0.91 = 0.09\), which is a good arithmetic check.
Step 7 — Find the mean for the unprepared search:
$$\mu = \frac{1}{p} = \frac{1}{0.70} \approx 1.43$$Step 8 — Flip the success to answer part g: Now the event that stops the survey is finding a resident who does have adequate supplies, so \(p = 0.30\).
$$\mu = \frac{1}{0.30} \approx 3.33$$Prepared residents are rarer, so the expected wait is more than twice as long.
Answer: a) \(X\) = the number of California residents surveyed until one is found without adequate earthquake supplies; b) 1, 2, 3, ...; c) \(X \sim G(0.70)\); d) 0.91; e) 0.09; f) about 1.43 residents; g) about 3.33 residents.
Key Terms
geometric experiment — a statistical experiment in which independent trials with a constant success probability are repeated until the first success occurs.
geometric random variable — the count produced by a geometric experiment: either the trial number of the first success (Case I) or the number of failures before it (Case II).
memoryless — the property that the probability of waiting \(n\) more trials for a success does not depend on how many failures have already happened.
common ratio — the constant factor \(r = 1 - p\) between consecutive probabilities in a geometric distribution.
\(X \sim G(p)\) — notation stating that \(X\) has a geometric distribution with success probability \(p\) on each trial.