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Solving Binomial Probability Problems

Discrete Mathematics · Chapter 8: Binomial Probability, Recurrence Relations, and Generating FunctionsSection 8.3 · lesson 48 of 878 min · free with a free account

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In this lesson

Exact, at-least and at-most probabilities, computed from the formula.

  • Restate and check the binomial model
  • Compute exact probabilities in full
  • Use the complement rule for at-least questions
  • Choose the cheaper direction of summation

Transcript

Everything said in the video, section by section (1,217 words). Read along, or search it.

Binomial Probability, Recurrence Relations, and Generating Functions

Solving Binomial Probability Problems

Section eight point two derived the binomial formula. This section is about applying it fluently. There are three question types. Exact, at least, and at most. The skill is recognizing which one is being asked, and computing it correctly and efficiently, straight from the formula.

The Opening Problem

When Do Too Many Pings Go Missing?

A ride-share platform's matching service depends on GPS pings from each driver's phone. In poor coverage, like a tunnel, a crowded stadium, or a rural gap, some pings are dropped in transit. We'll use an illustrative drop rate of ten percent. The algorithm sends ten pings per matching cycle. If too many go missing, it falls back to a stale, cached location. Before setting that threshold, an engineer needs a specific number. If each ping drops independently with probability zero point one, what's the probability that at least three of the ten are lost?

Definition 8.3.1

The Binomial Model, Restated

First, restate the model. X is binomial with parameters n and p if it counts the successes in n independent Bernoulli trials, each with success probability p. Its probability mass function gives the chance of exactly k successes. n choose k, times p to the k, times one minus p to the n minus k. Check the pings. There are ten per cycle, fixed. Each is delivered or dropped. And drops are treated as independent, at a constant rate of zero point one. So X, the number of dropped pings, is binomial with n equal to ten and p equal to zero point one.

3 Question Types

Exact, At Least, At Most

Three question types come up again and again. An exact question asks for precisely k successes. That's a single use of the formula. An at least question asks for k or more successes. That's a cumulative sum from k up to n. And an at most question asks for k or fewer. That's a cumulative sum from zero up to k. Telling them apart correctly is the main skill in this section.

Worked Example 8.3.1

Exactly 2 Dropped Pings

First, an exact probability. What's the chance that exactly two of the ten pings are dropped? The parameters are n equals ten, k equals two, and p equals zero point one, so one minus p is zero point nine. The binomial coefficient, ten choose two, is ten times nine over two, which is forty-five. The success power is zero point one squared, zero point zero one. The failure power is zero point nine to the eighth, about zero point four three zero five. Multiplying gives about zero point one nine three seven. Roughly a nineteen percent chance of exactly two dropped pings.

Worked Example 8.3.2

Exactly 1 False Positive in 20

The same formula transfers directly to a different domain. A payment platform's fraud screen flags a legitimate transaction with an illustrative probability of zero point zero two, independently. In a batch of twenty transactions, what's the probability of exactly one false positive? Twenty choose one is twenty. Zero point zero two to the first is zero point zero two. And zero point nine eight to the nineteenth is about zero point six eight one two. Multiplying gives about zero point two seven two five. Roughly a twenty-seven percent chance. That's not rare at all. Which is why fraud teams care more about the chance of two or more false positives, than about any single exact count. One note. The source text misprints that power of zero point nine eight. The recomputed value is used here, and it matches the figure the source uses later.

Theorem 8.3.1

At Least k - and Its Complement

At least k means k, k plus one, and so on, up to n. Written directly, that's a sum of many terms. But there's a second form. The probability of at least k equals one minus the probability of at most k minus one. That's not an approximation. The events at least k, and at most k minus one, split every possible value from zero to n into two disjoint, exhaustive groups. So their probabilities must add to one. It's a shortcut exactly when k minus one is small. Summing a few terms on one side beats summing many on the other.

Worked Example 8.3.3

At Least 3 of 10 Pings Lost

Now finish the opening problem. What's the probability that at least three of the ten pings are dropped? Summing directly from three to ten takes eight terms. The complement, from zero to two, takes only three. So use the complement. The probability of none dropped is zero point nine to the tenth, about zero point three four nine. Exactly one is about zero point three eight seven. And exactly two, from before, is about zero point one nine four. Those sum to about zero point nine three. So the probability of at least three is about zero point zero seven zero two. About seven percent of cycles would trigger the cached-location fallback. Whether that's acceptable is an engineering decision. But it can't be made responsibly without this number.

Figure 8.3.1

The Whole Distribution, and Its Tail

Here's the full distribution of dropped pings. Most of the probability sits at zero, one, and two. The gold bars, for three or more, are the tail that triggers the fallback. Together they add up to about zero point zero seven. Beyond six, the probabilities are so small they round to zero at four decimal places.

Theorem 8.3.2 and Worked Example 8.3.4

At Most 1 False Positive

At most k is the mirror image. It's a direct sum from zero up to k. When k itself is small, no complement is needed. The complement still applies in the other direction. It's worth using when k is close to n. Back to the fraud screen. The review team can absorb at most one false positive per batch, without a manual audit. What's the probability of that? k is one, close to zero, so sum directly. Two terms, instead of nineteen for the complement. No false positives has probability zero point nine eight to the twentieth, about zero point six six eight. Adding the zero point two seven two five from before gives about zero point nine four. So about ninety-four percent of batches need no audit. And about six percent do trigger one, a number the team can weigh directly against the analyst hours an audit costs.

Choosing a Computation Path

Which Direction Is Cheaper?

This table compares the three question types by the practical choice that matters when solving by hand. An exact question is always a single term. There's no direction to choose. For at least k, sum the few complement terms below k, when k is small. For at most k, sum directly when k is small. But when k is close to n, use the complement instead.

Takeaways

Recognize, Then Sum Efficiently

To summarize. The binomial formula applies only when the trials are fixed in number, two-outcome, and independent with a constant success probability. Verify that before computing. An exact question is one term. At least and at most questions are cumulative sums. The complement rule is a logical equivalence, not an approximation. Use it whenever it means summing fewer terms. And boundary errors are the most common mistake here. Here's something to think about. A common slip is writing the probability of at least three as one minus the probability of at most three, instead of at most two. Explain, in terms of which outcomes each event contains, why at least three and at most three aren't complements, while at least three and at most two are. Then consider more than k, strictly greater. Does it need the same shift against at most k? What does that say about the phrases at least, and more than?

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FAQ

Questions students ask us

About Solving Binomial Probability Problems

How do I check that a problem is binomial before using the formula?
Confirm that the number of trials is fixed, each trial has two outcomes, the trials are independent, and the success probability stays constant. Then X, the count of successes, is binomial with parameters n and p. For example, 10 pings per cycle, each delivered or dropped independently with probability 0.1, gives n=10 and p=0.1 for the number dropped.
How do I compute an exact binomial probability, like exactly 2 of 10 pings dropped?
Use one term of the formula, P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}. With n=10, k=2, p=0.1, this is \binom{10}{2}(0.1)^2(0.9)^8 = 45(0.01)(0.4305)\approx 0.1937. So there is about a 19% chance of exactly two dropped pings. An exact question never needs a sum.
How do I use the complement rule to find an "at least" probability, and when does it save work?
Use P(X\ge k)=1-P(X\le k-1), which is exact, not an approximation. It saves work when k is small. For at least 3 of 10 pings dropped, add three terms instead of eight: 1-(0.3487+0.3874+0.1937)\approx 0.0702. For "at most" questions, sum directly when k is small and use the complement when k is close to n.
Why is "at least 3" not the complement of "at most 3"?
The outcome X=3 belongs to both events, so they overlap and their probabilities add to more than 1. The complement of X\ge 3 is X\le 2, so P(X\ge 3)=1-P(X\le 2). Boundary errors like this are common. Likewise, "more than k" means X>k, whose complement is X\le k, so it needs no shift.

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