Coin-flip example
\(n=10, p=0.5, k=7 \rightarrow P(7) \approx 11.7\%\)
Binomial Distribution
Calculate binomial probabilities and cumulative probabilities
Explore the binomial distribution
Drag n (trials), p (success probability), and k (the outcome to highlight). The bar at k turns green; the rest of the PMF stays gray.
Quick presets
10
150
0.50
0.001.00
5
010
Predict what will happen
What does the bar chart look like when p ≈ 0.5 vs p ≈ 0.05 with the same n?
Switch between the 'Fair coin' and 'Rare event' presets and watch the shape.
Notice
The mean is np: at p = 0.5 with n = 10 the peak is at k = 5; at p = 0.7 with n = 20 the peak shifts to ≈ 14. Variance grows with n but is largest at p = 0.5.
Common mistake
P(X = k) is not the same as P(X ≤ k). The chart shows the PMF (exactly k); cumulative probabilities sum the bars from 0 up to k.
Why it works
Each trial is an independent Bernoulli draw. Multiplying p^k by (1−p)^(n−k) gives the probability of one specific sequence; the binomial coefficient counts how many sequences yield exactly k successes.
Concept check
Which of these scenarios actually fits the binomial model?
Results
Final Answer
\(P(X = 5) = 24.61\%\)
Step-by-step Solution
- Given: \(n = 10\), \(p = 0.50\), \(k = 5\)
- Binomial coefficient: \(\binom{10}{5} = \frac{10!}{5!\,(10-5)!} = 252\)
- Binomial PMF: \(P(X=k) = \binom{n}{k}\,p^k\,(1-p)^{n-k}\)
- Substitute and compute: \(P(X=5) = 252 \times 0.50^{5} \times 0.50^{5} = 0.246094\)
P(X ≤ 5)
62.30%
P(X ≥ 5)
62.30%
Mean / σ
μ = 5.00, σ = 1.58
Binomial distribution
The binomial distribution counts the number of successes in a fixed number n of independent Bernoulli trials, each with the same success probability p.
It applies when all four conditions hold:
- A fixed number of trials n.
- Each trial has only two outcomes (success / failure).
- All trials are independent.
- The probability p of success is the same for every trial.
Distribution parameters: \(\mu = np\), \(\sigma^2 = np(1-p)\), \(\sigma = \sqrt{np(1-p)}\)
\(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\)
Worked Examples
Quality-control example
\(n=20, p=0.1, k=2 \rightarrow P(2) \approx 28.5\%\)