Try
A synthetic process runs 24 independent trials. Each trial either succeeds or fails, with the same success probability 0.18. Let X count successes. Before calculating, write the event “at least six successes” as a list beginning 6, 7, 8 and ending at 24. Does a complement query need to include six, or stop at five?
This lesson uses a local calculator as one way to evaluate the probability. A suitable probability table, a verified sum or other technology can also evaluate it. The transferable skill is defining the random variable, model and event correctly, not memorizing one device’s menu. These probabilities concern a stipulated model, not a claim that a real process meets its assumptions.
Predict
The expected count is np=24·0.18=4.32. Predict whether at least six successes should be less likely than at least five, and whether more than six should equal at least six. Counts are discrete: exactly six has positive probability here, so changing a boundary can change the answer appreciably.
Write the complement before touching a control: the opposite of X≥6 is X≤5. “Not at least six” does not include six. This one-count difference is often more important than an extra displayed decimal.
Experiment
Enter n=24, p=0.18 and threshold 6. Select the at-least event and calculate. Record the event, inputs and result before changing anything. Then calculate the lower cumulative probability through five and subtract it from one independently.
| Query | Included outcomes | Worked result |
|---|
| P(X≤5) | 0 through 5 | about 0.7457662038 |
| P(X≥6) | 6 through 24 | about 0.2542337962 |
| P(X>6)=P(X≥7) | 7 through 24 | about 0.1256114205 |
The first two results sum to one apart from display rounding. The last is smaller because it excludes the mass at six. Compute the difference 0.2542337962−0.1256114205≈0.1286223757 and check it against the point event X=6. This is a second verification based on disjoint events, rather than repeating the same button press.
Now change only p to 0.30 and predict how the at-least-six probability changes before calculating. More likely successes make this upper-count event more likely under the same independent-trial construction. Reset to the worked parameters before comparing your saved handwritten values. The tool’s temporary state is not a graded response and does not award mastery.
Understand
For a binomial count, the point probability is P(X=j)=C(n,j)pʲ(1−p)ⁿ⁻ʲ, where C(n,j) counts the subsets of j successful trial positions. A lower CDF, F(k), adds those probabilities for j from zero through k. An at-least probability is 1−F(k−1); a strict upper probability is 1−F(k). The two formulas differ because X can equal k.
For an inclusive range a≤X≤b, subtract F(a−1) from F(b). Alternatively add point probabilities a through b. The sets must be disjoint before adding their probabilities directly. Do not add P(X≥a) to P(X≤b): their overlap would be counted twice.
The binomial model needs a fixed number of trials, a defined binary outcome, independent outcomes and the same success probability on each trial. Sampling without replacement from a small finite population violates literal within-sample independence. Repeated observations on one person can also be dependent. A calculator cannot diagnose these design problems from n and p alone. If collection stops when a target success count is reached, the number of trials becomes the random quantity, which is a different question.
Endpoint probabilities provide useful checks. When p=0, all probability is at X=0; when p=1, all is at X=n. These are valid binomial models, not numerical failures. An interior normal approximation is inappropriate at these endpoints, whereas the discrete probability calculation remains meaningful. Also, P(X≥0)=1 and P(X≤n)=1 for every admissible p.
The bounded tool accepts integer n from 1 through 200, p from 0 through 1 and integer threshold k from 0 through n. Values outside its interface are not impossible mathematical models; they are outside this teaching calculator’s contract. Keep extra digits during complements and sums, then round the final answer to the requested precision. A display rounded to zero does not automatically prove an exact model probability is zero.
Apply
Try a new calculation without copying the worked boundary: under independent trials with n=10 and p=0.5, find P(4≤X≤6). Use two lower-CDF queries and then a sum of three point queries. Explain what happens if someone subtracts F(4) instead of F(3).
Check the event and arithmetic after calculating
Use F(6)−F(3). For p=0.5, each ordered ten-trial success/failure sequence has probability 1/1024. The relevant counts are C(10,4)=210, C(10,5)=252 and C(10,6)=210. Thus the range probability is 672/1024=0.65625. Subtracting F(4) removes the 210/1024 probability at four and gives a different event. This exact count check is available because the example has equal-probability elementary sequences; it does not replace the general binomial formula for other p.
Continue
For unfamiliar practice, record four things: the trial model, the event in symbols, the calculator query and an independent check. Written feedback accepts stated numerical tolerances; it also asks whether the model and boundary are justified. A matching decimal without those decisions is not sufficient evidence of the calculation skill.
