The Binomial Probability Formula | Free Discrete Mathematics Lesson (4 College Credits) | UPI Study

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The Binomial Probability Formula

Discrete Mathematics · Chapter 8: Binomial Probability, Recurrence Relations, and Generating FunctionsSection 8.2 · lesson 47 of 878 min · free with a free account

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In this lesson

Why it has exactly the shape it does - one piece at a time.

  • Count the arrangements with exactly k successes
  • Find the probability of 1 specific sequence
  • Assemble the formula with the addition rule
  • Read what each factor contributes

Transcript

Everything said in the video, section by section (1,091 words). Read along, or search it.

Binomial Probability, Recurrence Relations, and Generating Functions

The Binomial Probability Formula

Section eight point one established when an experiment is binomial. This section derives the formula for binomial probabilities. It's one of the most heavily used formulas in discrete probability. And it rests on nothing more than counting, and the multiplication rule for independent events. The goal is to understand exactly why it has the shape it does, one piece at a time.

The Opening Problem

Clean Posts, Wrongly Flagged

An automated moderation system reviews social media posts with a trained classifier, flagging posts that appear to break platform policy. Suppose that on posts known to be fully compliant, it still wrongly flags them at a steady rate of about one in ten. We'll use p equal to zero point one as an illustrative figure. Each post is scored on its own, so whether one clean post gets flagged has no bearing on the next. If a batch of n compliant posts goes through, what's the probability that exactly k of them are wrongly flagged? Section eight point three applies the answer to realistic batch sizes. Here, the goal is the formula itself.

Definitions 8.2.1 and 8.2.2

Bernoulli Trials, Independently Repeated

Two definitions pin the experiment down. A Bernoulli trial has exactly two outcomes, success and failure, with a fixed probability of success, p. And trials are independent if no outcome affects the probability of any other. Then the probability that all of them happen is the product of their individual probabilities. Here, a success is a wrong flag, the outcome being counted, even though it's an error. Let X be the number of successes among n independent trials. X isn't a probability. It's a count, from zero to n. To keep every step explicit, use a small case. Three posts, p of zero point one, and a target of exactly two.

8.2.2 Counting the Arrangements

8 Sequences, 3 With 2 Successes

First, count the arrangements. Each of three trials resolves to success or failure, so there are two cubed, eight outcome sequences. Counting the successes in each, exactly three sequences have precisely two. Success, success, failure. Success, failure, success. And failure, success, success. That count also follows directly. Choosing which two of the three positions hold the successes determines the whole sequence. The number of ways to make that choice is three choose two, which is three. In general, a sequence of n trials with exactly k successes is fixed by choosing the k success positions. That's n choose k.

Worked Example 8.2.B

1 Specific Sequence

Now fix one specific sequence. Success, failure, success. Because the trials are independent, its probability is the product of the three trial probabilities. p, times one minus p, times p. That's p squared times one minus p. With p equal to zero point one, that's zero point one, times zero point nine, times zero point one. Zero point zero zero nine. Check the other two sequences with two successes. Each is also zero point zero zero nine. That's no coincidence. Any product of two factors of p and one factor of one minus p is the same, in whatever order the factors appear.

8.2.3 The Shared Probability

Every k-Success Sequence Weighs the Same

That generalizes directly. Any specific sequence of n independent trials, with k successes and n minus k failures, has probability p to the k, times one minus p to the n minus k. There are k factors of p, and n minus k factors of one minus p. And because multiplication doesn't depend on order, it's the same value for every one of the n choose k sequences, wherever the successes fall.

8.2.4 Assembling the Formula

Add the Sequences - They Can't Overlap

Now assemble the pieces. There are n choose k sequences with exactly k successes, and each has the same probability. The event X equals k is the union of all of those sequences. A single run of n trials produces exactly one outcome sequence. So two different sequences can't both occur. They're mutually exclusive. That means the addition rule applies. The probability of X equals k is the sum of n choose k identical terms. That gives the binomial probability formula. The probability of exactly k successes is n choose k, times p to the k, times one minus p to the n minus k. It holds when all four conditions from section eight point one are met. A fixed n, two outcomes per trial, a constant p, and independence.

Worked Example 8.2.C

Exactly 2 of 3 Posts Flagged

Back to the batch of three posts. The probability that exactly two are wrongly flagged is three choose two, times zero point one squared, times zero point nine. That's three times zero point zero one times zero point nine, which is zero point zero two seven. The whole derivation is a count of arrangements, times the shared probability of any one arrangement. And the addition rule justifies it, because the arrangements can't overlap.

Figure 8.2.1

The Whole Tree, n = 3, p = 0.1

This tree shows all eight possible sequences as paths. A success branch has probability zero point one, and a failure branch zero point nine. Each leaf's probability is the product of the branches on its path. The three highlighted leaves are the sequences with exactly two successes. Each is zero point zero zero nine. Adding them gives zero point zero two seven, matching the formula exactly.

Table 8.2.1

Anatomy of the Formula

This table breaks the formula into its three factors. n choose k counts the sequences with exactly k successes. It comes from counting arrangements. p to the k is the probability contributed by the successful trials, and one minus p to the n minus k by the failed ones. Both come from the multiplication rule for independent trials. Notice how each behaves if k and n minus k swap. n choose k stays the same. The other two factors generally don't. And n choose k is a count, not a probability. It only becomes part of one once it's multiplied by the shared probability.

Takeaways

Count, Multiply, Add

To summarize. Every sequence with exactly k successes among n independent trials has the same probability, wherever the successes fall. That probability is p to the k, times one minus p to the n minus k. n choose k counts those sequences. And the probability of exactly k successes is a sum of that many identical terms, valid because the sequences are mutually exclusive. The formula needs all four conditions at once. A fixed n, two outcomes, a constant p, and independence. Here's something to think about. Three choose one and three choose two are both three. So someone might expect the probability of one success to equal the probability of two. With p equal to zero point one, compute the probability of exactly one success, and compare it with zero point zero two seven. Are they equal? Which factor is responsible for any difference? And is there a special value of p that would make them match?

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FAQ

Questions students ask us

About The Binomial Probability Formula

What is the binomial probability formula and what does each part mean?
The probability of exactly k successes in n independent trials is P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}. The factor \binom{n}{k} counts the arrangements with exactly k successes, p^k covers the successful trials, and (1-p)^{n-k} covers the failed ones. It requires a fixed n, two outcomes per trial, a constant p, and independence.
Why does the binomial probability formula include n choose k?
The factor \binom{n}{k} counts how many different sequences contain exactly k successes, because choosing which k of the n positions hold the successes fixes the whole sequence. For n=3 and k=2, \binom{3}{2}=3 gives the sequences SSF, SFS, and FSS. Each has the same probability, so the count multiplies that shared value.
Why can you add the probabilities of the sequences with exactly k successes?
You can add them because one run of n trials produces exactly one outcome sequence, so two different sequences cannot both occur. They are mutually exclusive, and the addition rule applies. Every sequence has probability p^k(1-p)^{n-k}, so adding \binom{n}{k} identical terms is the same as multiplying that probability by \binom{n}{k}.
How do you find the probability of exactly 2 successes in 3 trials with p = 0.1?
The probability is 0.027, found from \binom{3}{2}(0.1)^2(0.9)^1 = 3(0.01)(0.9). Each of the three sequences with two successes, such as success, failure, success, has probability 0.1\cdot 0.9\cdot 0.1 = 0.009. Adding the three mutually exclusive sequences gives 3(0.009)=0.027, which matches the formula.

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