Binomial models and calculator tails

A count needs a trial model

Let X count defective components among five tested components. Before calculating, decide what one trial is and what counts as a success. In probability, “success” means the event being counted; it need not be desirable. Here one trial tests one component, and a success is finding a defect.

Write X∼ B(n,p) when all four conditions hold:

• there is a fixed number n of trials;
• each trial is classified as success or failure;
• the trials are independent;
• the success probability p is the same on every trial.

The parameters satisfy n = 0,1,2,… and 0 ≤ p ≤ 1. The possible counts run from 0 to n. At the boundary p = 0, X is always zero; at p = 1, X is always n. A binomial model can therefore include a count with no variability.

A fixed sample size does not establish the other conditions. Components made during the same machine fault may have related defect outcomes. Testing a large stable production run might support an independent constant-p approximation, but the manufacturing setting must justify it.

Count all orders for an exact result

For X∼ B(n,p) with 0 < p < 1,

The product p^r(1-p)^(n-r) is the probability of one particular order of r successes and n − r failures. The combination counts the different choices of success positions. Independence allows multiplication, and constant p makes all those orders have the same probability.

For five independent trials with p = 0.3, exactly two successes can occupy (5 choose 2) = 10 different pairs of positions. Thus

Using only (0.3)²(0.7)³ counts one specified order, such as success-success-failure-failure-failure. It misses the other nine orders.

The complete distribution in this small example provides a useful check:

These probabilities sum to one. They are not six equally likely outcomes: each count collects a different set of trial sequences.

Mean and variance describe repeated experiments

For a binomial count,

Use these results without needing their proofs. For B(5,0.3), the mean is 1.5, the variance is 1.05 and the standard deviation is square root of 1.05 ≈ 1.025.

A mean of 1.5 does not say that one experiment produces 1.5 successes; every observed count is an integer. Across many independent repetitions of the whole five-trial experiment, the average count tends towards 1.5. The standard deviation describes spread in counts; it is the square root of the variance.

The two formulas can also recover unknown parameters. If a binomial variable has mean 6 and variance 4.2, divide variance by the non-zero mean:

So p = 0.3, then n = 6/0.3 = 20. Check that n is an integer and p lies between zero and one before accepting the model. When the mean is zero, this division is unavailable; the count is always zero, and those summaries do not identify a unique n and p.

Translate a tail into allowed integers

A calculator’s point-probability command gives P(X = r). A cumulative binomial command commonly gives P(X ≤ r); check the tool’s own definition and the order of its n, p and boundary inputs. These two commands answer different questions.

For X∼ B(5,0.3):

For example, P(X ≥ 3) = 0.16308, while P(X > 3) = 0.03078. The difference is exactly P(X = 3) = 0.13230. Writing 1-P(X ≤ 3) for “at least three” wrongly removes the boundary count.

For an inclusive interval, subtract everything below its first allowed count:

Keep the model, integer event and probability expression in your working. A screen output alone does not show whether you entered the correct tail. Use stored precision in later calculations and round only the final requested answer.

Precision is part of the requested result. For example, P(X = 2) = 0.3087 is 0.309 to three significant figures, or 3.09 × 10⁻¹ in scientific notation. Writing 0.3090 states four significant figures. Three decimal places happens to give 0.309 here too, but the two requests differ for much smaller probabilities: 0.000622 has three significant figures and six decimal places.

Find a minimum trial count with its neighbouring integer

Suppose each independent trial succeeds with probability 0.2. To have probability at least 0.99 of one or more successes in n trials, require

Equivalently, 0.8ⁿ ≤ 0.01. Taking logarithms gives n ln 0.8 ≤ ln 0.01. Since ln 0.8 is negative, division reverses the inequality:

The least integer is 21. Check it in the original probability: n = 20 gives about 0.988471, while n = 21 gives about 0.990777. This also verifies a calculator’s integer search and prevents rounding 20.638 down to 20. The probability increases with n because 0.8ⁿ decreases.