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Applied Binomial Experiment Problems

Discrete Mathematics · Chapter 8: Binomial Probability, Recurrence Relations, and Generating FunctionsSection 8.7 · lesson 52 of 878 min · free with a free account

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In this lesson

Case studies that use Sections 8.1 to 8.6 side by side.

  • Verify the binomial conditions for a real scenario
  • Combine exact and cumulative probabilities
  • Summarize with the mean and standard deviation
  • Follow a 5-step checklist to a decision

Transcript

Everything said in the video, section by section (1,167 words). Read along, or search it.

Binomial Probability, Recurrence Relations, and Generating Functions

Applied Binomial Experiment Problems

This section brings together everything from sections eight point one through eight point six. No new formulas appear. Instead, four case studies apply the whole binomial toolkit side by side, the way a real capacity-planning decision needs it.

The Opening Problem

250 Sensors, Every Hour

A utility company's grid-monitoring team relies on two hundred fifty smart-grid sensors, each reporting a reading every hour. Each sensor independently fails to report in a given hour. We'll use an illustrative probability of zero point zero four. That one sentence raises several questions. Is it binomial? What's the chance of some specific number of dropouts? Of crossing a threshold? What's the average? And how much does it vary? In practice, none of these gets asked in isolation. A real decision needs all of them answered together, for the same scenario.

Recall

The Binomial Toolkit

Here's the toolkit. The conditions. A fixed number of trials, two outcomes each, and independent trials with a constant success probability. The probability mass function, n choose k, times p to the k, times one minus p to the n minus k. Cumulative probabilities, using the complement rule, or a table, whichever needs fewer terms. The mean, n p. And the variance, n p times one minus p, with the standard deviation its square root.

Case Study 8.7.1

Sensor Dropouts and Alert Thresholds

Start with the sensors. There are two hundred fifty, a fixed number. Each reports or doesn't. And dropouts are treated as independent, at a constant rate. So the count of hourly dropouts is binomial, with n two hundred fifty and p zero point zero four. What's the chance of exactly eight dropouts in an hour? The formula gives about zero point one one three five. Roughly an eleven percent chance in any hour. Grid policy treats fifteen or more simultaneous dropouts as degraded mode. Fifteen is far from zero, so use the complement. The chance of at most fourteen is about zero point nine two one. So fifteen or more has probability about zero point zero seven nine. About seven point nine percent of hours. The mean is two hundred fifty times zero point zero four, which is ten dropouts an hour. The variance is ten times zero point nine six, which is nine point six. So the standard deviation is about three point one.

Figure 8.7.1

Center, Band and Tail

Here's the full distribution. The gold bars sit inside the band from the mean minus one standard deviation, to the mean plus one. That's about six point nine to thirteen point one. Most of the probability sits inside that band. That's what typical spread means in practice. The fifteen-or-more alert region lies out in the right tail, about one point six standard deviations above the mean. It fires in roughly one hour out of every twelve or thirteen. Whether that alert rate is workable is an operations decision. These numbers are what make it an informed one.

Case Study 8.7.2

Why Mining Rewards Swing So Much

A different kind of question can be settled with the same tools. A cryptocurrency mining rig makes twenty thousand independent hash attempts in a work interval. Under a hypothetical, low-difficulty target, used here for illustration, each attempt produces a valid share with probability zero point zero zero zero two. The mean is twenty thousand times zero point zero zero zero two, which is four shares. The variance is just under four, so the standard deviation is about two. And the chance of finding no shares at all, is zero point nine nine nine eight to the twenty thousandth. That's about zero point zero one eight, or one point eight percent of intervals, despite averaging four. A standard deviation half the size of the mean is large. An individual miner's count swings a lot, purely from randomness. That's one mathematical reason, though not the only one, why individual miners often join pools that combine many participants' attempts.

Case Study 8.7.3

Two Thresholds, One Distribution

Sometimes one scenario needs both cumulative directions at once. During a flash sale, a payment gateway processes eight hundred checkouts a minute at peak. Each fails from a gateway timeout with an illustrative probability of zero point zero one five. The mean is twelve failures a minute, the variance eleven point eight two, and the standard deviation about three point four. Engineers treat five or fewer failures as routine. Five is small, so sum directly. The probability is about zero point zero two. Only about two percent of peak minutes are that calm. And twenty or more triggers a failover to a backup gateway. Twenty is far from zero, so use the complement. That's also about two percent of minutes. The two thresholds sit at roughly similar distances from the mean, in opposite directions, and land on similar probabilities. But neither could have been read off the mean and standard deviation alone.

8.7.4 A Checklist

5 Steps for Any Applied Question

All three case studies followed the same moves. Making them explicit gives a reusable checklist. Step one, verify the binomial conditions. Never skip it. Step two, locate the question. Is it one exact count, or a range? Step three, for a range, choose a direction. Is the threshold closer to zero, or to n? Step four, summarize with the mean and standard deviation, when a general profile is wanted. And step five, interpret. What decision does this number support, and for whom? An unlabeled number rarely answers the original question.

Case Study 8.7.4

The Checklist on Wafer Yield

Here's the checklist on a new scenario. A semiconductor line tests forty chips per wafer. Each is independently defective with probability zero point zero five. A wafer ships without rework if at most two are defective. Verify. The conditions hold. Locate. At most two is a range, so it needs a cumulative sum. Choose a direction. Two is close to zero, so sum directly. The probabilities of zero, one, and two defects add to about zero point six seven seven. Summarize. The mean is forty times zero point zero five, which is two. The variance is one point nine, so the standard deviation is about one point four. Interpret. About sixty-eight percent of wafers ship without rework. They average two defective chips, varying by about one and a half, which is exactly what a yield-improvement team would track over time.

Takeaways

Verify, Locate, Choose, Summarize, Interpret

To summarize. Applied binomial problems rarely need just one tool. A complete decision verifies the conditions, computes exact or cumulative probabilities, and summarizes with the mean and standard deviation. The direction of a cumulative sum is chosen the same way every time. Sum toward whichever end the threshold is nearer. The mean and standard deviation describe center and spread. But they don't substitute for a probability when a specific threshold decision is on the table. Here's something to think about. A colleague argues that once you know the mean of twelve, and the standard deviation of about three point four, the cumulative sums are redundant. Using the checkout numbers, explain what the probabilities of five or fewer, and twenty or more, tell an engineer that the mean and standard deviation alone don't. Then reverse it. Is there a decision in that case study, that the mean and standard deviation could answer perfectly well on their own? What distinguishes the two kinds of questions?

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FAQ

Questions students ask us

About Applied Binomial Experiment Problems

How do you verify that a real-world scenario is a binomial experiment?
Check that there is a fixed number of trials, two outcomes per trial, independent trials, and a constant success probability. For 250 sensors that each independently fail to report in an hour with probability 0.04, all four conditions hold. The hourly dropout count is then binomial with n=250 and p=0.04. Strongly linked failures would break independence.
When should you use the complement rule instead of adding terms for a cumulative binomial probability?
Use the complement when the threshold is far from zero, and add terms directly when it is near zero. For n=250 and p=0.04, finding P(X\ge 15) directly would take many terms. Instead, P(X\ge 15)=1-P(X\le 14)\approx 1-0.921=0.079, so about 7.9% of hours reach degraded mode.
Why can't the mean and standard deviation replace cumulative binomial probabilities?
The mean and standard deviation describe center and spread, but they do not give the chance of crossing a specific threshold. For 800 checkouts with p=0.015, the mean is 12 and the standard deviation is about 3.4. Only the cumulative sums show that P(X\le 5) and P(X\ge 20) are each about 0.02, which is what a failover decision needs.
What are the five steps for solving an applied binomial problem?
The steps are verify the binomial conditions, locate the question as an exact count or a range, choose the cumulative direction, summarize with the mean and standard deviation, and interpret. For 40 chips with p=0.05, P(X\le 2)\approx 0.677 by direct summing. The mean is 2 and the standard deviation is about 1.4. So about 68% of wafers ship without rework.

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