binompdf & binomcdf Calculator
Evaluate the TI-84 binomial functions online. binompdf(n, p, x) gives the probability of exactly x successes; binomcdf(n, p, x) gives the probability of at most x. This page also returns the "at least", "fewer than", and "more than" versions so you never have to work out the complement by hand.
Part of the TI-84 statistics functions guide, which shows when to use binompdf and binomcdf and every other DISTR-menu function.
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geometpdf & geometcdf Calculator
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Binomial vs Poisson vs Hypergeometric
Three questions decide the right counting distribution (fixed trials, independence, replacement) with one inspection scenario computed all three ways.
TI-84 Statistics Functions
Every TI-84 statistics function: DISTR (normalcdf, invNorm, tcdf, binompdf…) and STAT TESTS (T-Test, TInterval, 2-SampFTest, LinRegTTest…), with syntax, a decision guide, and an online calculator for each.
binompdf vs. binomcdf
Both functions describe a binomial experiment: n independent trials, each a success with the same probability p. The difference is what they add up:
binompdf(n, p, x) = C(n, x) · pˣ · (1 − p)ⁿ⁻ˣ = P(X = x)
binomcdf(n, p, x) = Σ binompdf(n, p, k) for k = 0…x = P(X ≤ x)
"pdf" stands for probability density (strictly, mass) function: one bar of the distribution. "cdf" is the cumulative distribution function: every bar from 0 up to x.
Translating Word Problems
| Phrase | Probability | TI-84 entry |
|---|---|---|
| exactly 3 | P(X = 3) | binompdf(n, p, 3) |
| at most 3 / no more than 3 | P(X ≤ 3) | binomcdf(n, p, 3) |
| fewer than 3 | P(X ≤ 2) | binomcdf(n, p, 2) |
| at least 3 | 1 − P(X ≤ 2) | 1 − binomcdf(n, p, 2) |
| more than 3 | 1 − P(X ≤ 3) | 1 − binomcdf(n, p, 3) |
| between 3 and 6 | P(X ≤ 6) − P(X ≤ 2) | binomcdf(n,p,6) − binomcdf(n,p,2) |
The most common mistake is the off-by-one in "at least": P(X ≥ 3) subtracts binomcdf up to 2, not 3, because 3 itself must stay in the answer.
Worked Example: Three-Point Shots
A player makes 30% of three-point attempts and shoots 10 times. What is the probability of making exactly 3, and of making at most 3?
- binompdf(10, 0.3, 3) = C(10, 3) × 0.3³ × 0.7⁷ = 120 × 0.027 × 0.0823543 = 0.266828.
- binomcdf(10, 0.3, 3) = P(0) + P(1) + P(2) + P(3) = 0.028248 + 0.121061 + 0.233474 + 0.266828 = 0.649611.
- At least 4 = 1 − 0.649611 = 0.350389.
Enter n = 10, p = 0.3, x = 3 above to reproduce every number, or explore the same experiment with the binomial distribution calculator.
Where to Find binompdf on a TI-84
- Press 2nd then VARS (DISTR).
- Scroll to A:binompdf( or B:binomcdf(.
- Enter trials, p, and x value, then select Paste and press ENTER.
Leaving the x value blank returns the whole list of probabilities, which is what the table above shows. In Excel use =BINOM.DIST(x, n, p, FALSE) for binompdf and =BINOM.DIST(x, n, p, TRUE) for binomcdf.
Related Guides and Calculators
The binomial distribution calculator gives the same probabilities as binompdf and binomcdf, with the mean, the variance and a chart. The coin flip probability calculator applies it to heads and tails, and the at least one probability calculator gives the chance of one or more successes. If you draw without replacement use the hypergeometric distribution, and for rare events over an interval the Poisson distribution. Read binomial vs Poisson vs hypergeometric to choose the model and TI-84 statistics functions for the other calculator commands.
Frequently Asked Questions
When should I use binompdf instead of binomcdf?
Use binompdf when the question asks for exactly one count of successes. Use binomcdf when it asks for a range that starts at 0, 'at most', 'no more than', or 'fewer than'. Ranges that start above 0 are built from binomcdf with subtraction.
How do I calculate 'at least' with binomcdf?
P(X ≥ x) = 1 − binomcdf(n, p, x − 1). For example, at least 4 successes in 10 trials with p = 0.3 is 1 − binomcdf(10, 0.3, 3) = 0.350389.
What conditions must hold to use binompdf?
A fixed number of trials n, only two outcomes per trial, the same success probability p on every trial, and independent trials. Sampling without replacement from a small population breaks independence: use the hypergeometric distribution instead.
Why does binompdf return a number like 3.2E-5?
That is scientific notation for 0.000032. Very small probabilities appear when x is far from the expected count n × p.
What is the mean of a binomial distribution?
The mean is n × p and the standard deviation is √(n × p × (1 − p)). For n = 10 and p = 0.3 the mean is 3 successes and the standard deviation is about 1.449.
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