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Grade 12 · Binomial probability

Binomial probability: Many moderately likely successes

Binomial probability: investigate many moderately likely successes with trial count n = 12; success probability p = 0.65.

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Watch the relationship

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Binomial probability: Many moderately likely successes. Success count: 0. Exact mass P(X=k): 0. Cumulative P(X≤k): 0. Mean np: 7.8Exact masses; no random sampling00.240612probabilityBars are probability masses at whole counts
There are exactly n independent Bernoulli trials with a fixed p. Bars show probabilities, not random outcomes or real observations. The linked Grade 12 lesson develops comparison of observed categorical counts with model expectations.

Starts paused. Play once, pause anywhere, or use Step to inspect the mathematics. Playback stops when this panel leaves the screen.

Make it your experiment

Change one value. Notice what follows.

The controls adjust the model. Numbers below describe the current frame. Decimals are rounded.

Success count
0
Exact mass P(X=k)
0
Cumulative P(X≤k)
0
Mean np
7.8

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Understand what you are seeing

The idea behind the motion.

Counting successes in a fixed set of independent, identical-probability trials gives a binomial model. Its probability masses belong to integer outcomes, while the expected count may be fractional. Scanning exact masses distinguishes a theoretical distribution from the noisy frequencies of one simulated sample. This investigation starts with Trial count n = 12; Success probability p = 0.65. Predict the result before playing, then change one parameter while keeping the other fixed to test the reason for the change.

A relationship to keep

P(X=k)=C(n,k)pᵏ(1−p)ⁿ⁻ᵏ; E[X]=np

Read the symbols alongside the explanation. A diagram shows the relationship; the assumptions tell you when it applies.

  1. STEP 1

    Set up the mathematical model

    Specify the number of trials and common probability, then check that the model's independence assumption is meaningful. The starting case is “Many moderately likely successes.”

  2. STEP 2

    Follow the changing quantity

    Scan counts from zero to n. Each highlighted bar reports exact mass and the cumulative probability through that count.

  3. STEP 3

    Explain and test the result

    Compare the distribution with its mean np and predict how an observed count table could be checked against expected counts. A good fit would not itself prove independence.

Your turn to explain

Make a prediction. Test your reasoning.

Keep Trial count n = 12; Success probability p = 0.65. Pause the timeline at 60%. Given success count = 7, calculate exact mass p(x=k), cumulative p(x≤k), mean np. Show the substitution into the displayed formula.

Use the values specified in the question. Reset restores the initial values for this investigation.

Compare your explanation

For k=7, C(12,7)(0.65)^7(1−0.65)^5=0.20392. Adding this formula from k=0 through 7 gives 0.416655; the mean is 12(0.65)=7.8. Results: Exact mass P(X=k): 0.204; Cumulative P(X≤k): 0.417; Mean np: 7.8. Decimal values are rounded; retain the original parameters when checking.

Work through a full lesson

Connect the animation to a worked example and practice questions.

Grade 12 · Statistics

Compare observed categorical counts with a specified model

Compute a goodness-of-fit statistic from expected counts.