Grade 12 · Binomial probability
Binomial probability: A broad off-center count model
Binomial probability: investigate a broad off-center count model with trial count n = 15; success probability p = 0.4.
Predict what will happen, press Play, then pause and explain what changed. Every control also works without playback.
Watch the relationship
PausedHD animation studio
From experiment to screen.
Present a crisp Canvas scene, save a full-HD image, or capture your model as a silent video.
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 = 15; Success probability p = 0.4. 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.
- 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 “A broad off-center count model.”
- 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.
- 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 = 15; Success probability p = 0.4. Pause the timeline at 100%. Given success count = 15, 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=15, C(15,15)(0.4)^15(1−0.4)^0=0.000001. Adding this formula from k=0 through 15 gives 1; the mean is 15(0.4)=6. Results: Exact mass P(X=k): 0; Cumulative P(X≤k): 1; Mean np: 6. 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.