Binomial distribution

What is it about?

A Bernoulli chain of length $n$ repeats an experiment with only two outcomes (success or failure) $n$ times independently, always with the same probability of success $p$. The number of successes $X$ is then binomially distributed, $X \sim B(n;\,p)$:

With the cumulative distribution function $F(k) = P(X \le k)$:

• at most $k$: $P(X \le k) = F(k)$
• at least $k$: $P(X \ge k) = 1 - F(k - 1)$
• between $a$ and $b$: $P(a \le X \le b) = F(b) - F(a - 1)$

Expected value $\mu = n \cdot p$, standard deviation $\sigma = \sqrt{n \cdot p \cdot (1-p)}$. The interval $[\mu - 2\sigma;\, \mu + 2\sigma]$ usually holds about $95\,\%$ of the probability.

Try it

1. Drag the bound $k$ on the histogram and watch the calculation. Load example
2. Change $p$ and $n$: where does $\mu$ go, how wide is the histogram? B(20; 0.3)
3. Switch to the cumulative view: every bar becomes a step of $F(k)$. Step function
4. Run the experiment 1000 times and compare the orange outline with the bars.

Tasks

Task 1: Guessing in a test

$20$ questions with four answers each, guessing at random. Find $P(X = 5)$ and $P(X \ge 10)$. Load task

Task 2: 100 coin tosses

Find $\mu$, $\sigma$ and $P(40 \le X \le 60)$. Load task

Task 3: At least one

A part is faulty with probability $0.05$. Among $50$ parts, find $P(X \ge 1)$. Load task

Ideas for the classroom

• Use it after the Galton board: there the distribution emerges, here you calculate with it.
• Check the requirements together first (two outcomes, independent, same $p$) – drawing without replacement is a good counterexample.
• Common stumbling blocks:
  • “At least $k$” computed as $1 - F(k)$ instead of $1 - F(k - 1)$.
  • $P(a \le X \le b)$ computed as $F(b) - F(a)$, which misses $P(X = a)$.
  • Confusing $\sigma$ with the variance $n \cdot p \cdot (1-p)$.