Grade 12 · Binomial probability
Binomial probability: Eight fair trials
Binomial probability: investigate eight fair trials with trial count n = 8; success probability p = 0.5.
Predict what will happen, press Play, then pause and explain what changed. Every control also works without playback.
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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 = 8; Success probability p = 0.5. 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 “Eight fair trials.”
- 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 = 8; Success probability p = 0.5. Pause the timeline at 40%. Given success count = 3, 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=3, C(8,3)(0.5)^3(1−0.5)^5=0.21875. Adding this formula from k=0 through 3 gives 0.363281; the mean is 8(0.5)=4. Results: Exact mass P(X=k): 0.219; Cumulative P(X≤k): 0.363; Mean np: 4. Decimal values are rounded; retain the original parameters when checking.
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