2.10 Binomial Counts: Model, Calculate and Interpret

Counting Successes in a Fixed Set of Trials

A grower tests five seeds under a stated model: each seed independently germinates with probability 0.80. Let X count germinated seeds. Each trial is a seed; “success” means germination, not something universally good. X counts successes in a fixed number of independent trials with two outcomes per trial and the same success probability. These conditions make X a binomial random variable with $n=5$ and $p=0.80$.

“Failure” has probability $1-p=0.20$. The possible counts are 0 through 5. The physical situation must support the model: different growing conditions or a shared disease could undermine a constant probability or independence. Merely counting successes does not establish a binomial model.

Sampling five items without replacement from a tiny lot is a different situation. Once one success is removed, the chance on the next draw changes, so exact independence and constant p fail. Counting how many trials are needed until the first success also differs: the trial count is not fixed in advance. Always identify X and verify the design before reaching for a binomial formula.

Worked Example: Typical Five-Seed Counts

For a binomial count,

Over many independent sets of five seeds under this model, the average number germinated approaches four seeds per set. Counts typically deviate about 0.894 seeds from that mean. The variance is 0.80 square seeds; it is not the SD. These parameters describe five-seed counts, not individual seed heights or a guarantee that exactly four will germinate.

Why the Binomial Coefficient Appears

One particular sequence with four germinations and one failure has probability $0.8^4(0.2)$. The failure could occur in any of five positions, so the event “exactly four” contains five equally probable sequences. In general,

The coefficient counts ways to choose the x success positions. The powers account for x successes and n−x failures. For X=4,

Thus 40.96% of repeated five-seed sets would have exactly four germinations in the long run. The complete distribution is useful for checking endpoint language:

x	0	1	2	3	4	5
P(X=x)	0.00032	0.00640	0.05120	0.20480	0.40960	0.32768

All six values are possible and their probabilities sum to 1. “At least four” includes 4 and 5, giving 0.73728. “More than four” includes only 5, giving 0.32768. “Between two and four, inclusive” adds the entries for 2, 3 and 4, giving 0.66560. Technology can evaluate these sums or cumulative probabilities, but write the event first: $P(X\ge4)=1-P(X\le3)$, not $1-P(X\le4)$.

An Empirical Distribution Is Also Useful

To simulate one five-seed set, generate five independent uniform digits 0–9; 0–7 mean germination and 8–9 mean failure. Record the success count X once for that complete set. Repeat for many sets. A frequency table for X=0,1,2,3,4,5, divided by the number of sets, estimates the probability distribution. Add the estimated entries belonging to the requested event.

An empirical probability can differ from the exact value through finite-run variation. A zero count at a rare possible value does not prove the theoretical probability is zero. Keep n, p, success definition, digit encoding and event endpoints consistent between the exact and simulated approaches. The exact table above is constructed by probability rules; a recorded simulation supplies an approximation to the same model. Neither a simulator’s availability nor a list of settings is a completed run—execution requires recorded results and a matching count calculation.

Try It Yourself

The following original practice applies the ideas above to a fresh setting. Show the reasoning requested in each part. Answers and task-specific rubrics are on the separate solutions page.

Original generated practice for the Fall 2026 course.

Which Defect Counts Are Binomial?

Setting I: inspect 20 independently produced components; a stated stable production model gives each component defect probability .02. X counts defective components. Setting II: a fixed box contains 40 components, exactly 12 defective; select 20 without replacement and let Y count defects. Setting III: inspect independent components with defect probability .02 until the third defect appears; Z is the number inspected.

Part a

Justify whether X is binomial by identifying its numerical success count, fixed n, two outcomes, success/failure probabilities, and independence/constant-probability assumptions. State its parameters.

Your reasoning

Part b

Justify why Y and Z do not have the same binomial structure. For each, identify the specific failed condition rather than merely saying “not binomial.”

Your reasoning

Worked answer and rubric

Mean, Spread and a Coupon Probability

Under a supplied model, each mailed coupon is redeemed independently with probability .20. One campaign sends 25 coupons to distinct households. X counts redemptions in a campaign; repeated campaigns follow the same model. The model probability of at least eight redemptions is .1091228.

Part a

Calculate the binomial mean and standard deviation with substituted formulas. Also state the variance to make clear which quantity requires a square root.

Your reasoning

Part b

Interpret the mean, SD and .1091228 probability in the campaign population. Explain why a mean of five does not mean every campaign gets five redemptions or that a single coupon has probability five.

Your reasoning

Worked answer and rubric

Estimate a Binomial Event from Simulation

Each arcade play independently awards a token prize with probability .25 under a specified model. One trial consists of four plays, and X counts prizes. The simulation generated four independent uniform digits 0–3 for each trial; 0 means prize. The complete 100-trial count table is shown.

Complete results of 100 newly generated four-play trials

X: token prizes	0	1	2	3	4
Number of trials	27	45	21	7	0

Part a

Construct the empirical probability distribution for all X values. Use it to estimate P(X≥2), showing which entries you add and interpreting the result.

Your reasoning

Part b

Independently run 50 new four-play trials using an equivalent valid random device. Record every four-result trial and X, construct the new count/probability table and estimate P(X≥2). Explain why your estimate may differ from .28 without changing the model.

Your reasoning

Part c

Construct the exact binomial probability distribution by rules for n=4,p=.25. Calculate the exact P(X≥2), compare it with .28 and identify which is exact and which is estimated.

Your reasoning

Worked answer and rubric

Exact Binomial Boundaries

A service company’s supplied model gives each repair visit independently a .30 chance of needing an extra part. A route has eight visits. Let X count visits needing an extra part; thus X is binomial with n=8,p=.30.

Part a

Write the complete binomial probability function with its admissible x values. Calculate P(X=3), showing the combination factor and both powers. Interpret the result.

Your reasoning

Part b

Calculate P(X≥3) and P(X>3) with explicitly identified sums or complementary cumulative expressions. Explain why their difference is P(X=3).

Your reasoning

Part c

Calculate P(2≤X≤4), P(X=0) and P(X=8). Show the included values/formulas and check that every answer is a valid probability.

Your reasoning

Worked answer and rubric