--- title: "The Binomial Distribution" description: "Count successes in a fixed number of independent trials that each have the same probability. The number of arrangements times the probability of one arrangement gives each term." canonical: https://lightmysky.com/learn/mathematics/the-binomial-distribution-mt_X2kVGK67Md source: https://lightmysky.com/learn/mathematics/the-binomial-distribution-mt_X2kVGK67Md.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # The Binomial Distribution Count successes in a fixed number of independent trials that each have the same probability. The number of arrangements times the probability of one arrangement gives each term. Subject: Mathematics · Area: Probability · Ages 16 to 17 Page: https://lightmysky.com/learn/mathematics/the-binomial-distribution-mt_X2kVGK67Md ## Ready when they can - Check the four conditions before choosing a binomial model - Calculate probabilities for an exact success count and related events - Explain dependence in multiple draws without replacement from a mixed finite population ## Lesson: Counting the wins in a fixed number of goes You used the product rule to count lunches and distributions to describe payouts. Now let X count wins in a book of 10 coconut-shy goes. Assume the results are independent and every go has win probability 0.2. A binomial model describes this count. It needs fixed trials, success or not on each trial, constant success probability and independence. Knowing the success count beforehand is not a condition. *(drawing: W means win and M means miss. This is one ordered result with three wins among ten goes.)* Check the four conditions separately. Trials may have several physical outcomes grouped as success or not, such as six or not-six on a die. Resetting a stall does not prove constant probability or independence: learning, fatigue or different players may change its chances. A calculation can return a number when conditions fail. Check whether it is an exact model or a useful approximation. For n independent trials, let p be the success probability and r the success count. One ordered result has probability p^r × (1-p)^(n-r). In our book, WWWMMMMMMM has probability 0.2³ × 0.8⁷ = 0.0016777216. Every order with three wins uses the same factors, so it has the same probability. Independence allows multiplication; constant p makes the factors match. *(drawing: This gives one arrangement's probability. Exactly three wins can occur in many arrangements.)* **Example.** Choose the three winning positions. Ordered selections give 10 × 9 × 8 = 720. Each set appears in 3 × 2 × 1 = 6 orders, so there are 720/6 = 120 arrangements. For any success count, divide ordered selections by the orderings of the selected positions. Choosing successes also fixes failures. The arrangements cannot occur together, so add their equal probabilities: P(X = 3) = 120 × 0.0016777216 = 0.201326592. A drum holds 20 tickets, including 5 winners. Draw three uniformly without replacement. The second win chance is 4/19 after a win but 5/19 after a miss, showing dependence. Before seeing earlier results, each draw still has marginal win probability 5/20. Returning each ticket and mixing thoroughly before a fresh uniform draw gives independent draws with the same win probability. Write X ~ B(n,p) for this binomial count. C(n,r), read 'n choose r', counts sets of r positions among n. Then P(X = r) = C(n,r) × p^r × (1-p)^(n-r). C(n,0) = C(n,n) = 1. Multiply the count by one arrangement's probability; adding the count to a probability does not total the alternatives. Define success first: when counting misses, p must be the miss probability. Equal marginal probabilities do not prove independence. Toss one fair coin and copy its result into six positions. Each entry has head probability 0.5, but learning one entry reveals every other entry. The head count is only 0 or 6, unlike B(6,0.5). Independence requires earlier results to leave the probabilities of later results unchanged. There is one all-success arrangement and one no-success arrangement. Their probabilities are p^n and (1-p)^n. At least one success is the complement of no successes, so P(X ≥ 1) = 1 - (1-p)^n. With n = 1, the complete distribution is P(X = 0) = 1-p and P(X = 1) = p. Arrangement counts are symmetric because selecting successes also selects the remaining failures. Consecutive counts have ratio C(n,r+1)/C(n,r) = (n-r)/(r+1), so they rise while this ratio exceeds 1. For even n, the middle count r = n/2 is largest; for odd n, the two middle counts tie. This is about arrangement counts, not necessarily the most likely count when p differs from 0.5. Multiple uniform draws without replacement from a mixed finite population are dependent. A binomial can approximate the count when the sample is small relative to the population; check accuracy for the event needed. Exceptions include a single draw, which gives B(1,p), and all-success or all-nonsuccess populations, whose counts match B(n,1) or B(n,0). **Recap.** Check fixed trials, two outcome categories, constant p and independence. Multiply the arrangement count by p^r(1-p)^(n-r). Draws without replacement from a mixed population are dependent. *Lesson, practice questions and flashcards written for Marble.* ## Practice 30 questions on this page, each with its working shown. ## Needs first - [Counting Outcomes with the Product Rule](https://lightmysky.com/learn/mathematics/counting-outcomes-with-the-product-rule-mt_74gSbYIG8p) - [Expected Value and Variance of a Discrete Random Variable](https://lightmysky.com/learn/mathematics/expected-value-and-variance-of-a-discrete-random-variable-mt_PFdo0Nm3FP) ## Optional connections - [The Binomial Expansion](https://lightmysky.com/learn/mathematics/the-binomial-expansion-mt_z16HZkNZ8c) ## Opens up - [Discrete Models Beyond the Binomial](https://lightmysky.com/learn/mathematics/discrete-models-beyond-the-binomial-mt_9zc5xYjDg_) - [Likelihood and the Maximum Likelihood Estimate](https://lightmysky.com/learn/computing/likelihood-and-the-maximum-likelihood-estimate-mt_crcBnSqkTE) - [The Normal Distribution and Standardised Scores](https://lightmysky.com/learn/mathematics/the-normal-distribution-and-standardised-scores-mt_Cs32DT9KnA) - [Testing a Binomial Proportion](https://lightmysky.com/learn/mathematics/testing-a-binomial-proportion-mt_ZJ8MTLYYKE)