Bernoulli & Binomial

A single coin flip and the count of successes across independent trials.

1. The idea

A Bernoulli $(p)$ random variable takes value $1$ with probability $p$ and $0$ with probability $1 - p$ . It models one yes/no trial: a coin flip, a trade filling, a default event.

Summing $n$ independent Bernoulli $(p)$ trials gives a Binomial $(n, p)$ random variable, the number of successes out of $n$ tries. Its distribution is $P (X = k) = (k n) p^{k} (1 - p)^{n - k}$ .

Because a Binomial is a sum of independent identical pieces, its mean and variance follow immediately from linearity: $E [X] = n p$ and $Var (X) = n p (1 - p)$ , no summation needed.

2. Why it matters

Binomial reasoning appears whenever a question involves counting successes in independent trials — win/loss streaks, order fills, A/B tests — and recognizing it turns a combinatorics problem into a one-line moment calculation.

3. Core mathematics

Parameters

n \in Z^{+}, p \in [0, 1]

$n$ trials, success probability $p$

Support

k \in {0, 1, \dots, n}

PMF

P (X = k) = (k n) p^{k} (1 - p)^{n - k}

Expectation

E [X] = n p

Variance

Var (X) = n p (1 - p)

Key property

X = i = 1 \sum n B_{i}, B_{i} \sim ii d Bernoulli (p)

sum of independent identical indicators

4. How to recognize it

Think of this when…

A fixed number of independent trials, each with the same success probability
Question asks for the probability of exactly $k$ successes, or the expected number of successes
Phrases like "flip a coin $n$ times", " $n$ independent trades", "each with probability $p$ "

5. Worked example

Worked example

A market maker's quote is filled independently with probability $0.3$ on each of $10$ attempts. Find the expected number of fills and the probability of exactly $2$ fills.

1
Model each attempt as Bernoulli $(0.3)$ , so the total fills $X \sim Binomial (10, 0.3)$ .
2
Expectation by linearity, no PMF needed.
$E [X] = n p = 10 (0.3) = 3$
3
Probability of exactly $2$ fills uses the PMF directly.
$P (X = 2) = (2 10) (0.3)^{2} (0.7)^{8}$
4
Compute the value.
$(2 10) = 45, 45 \times 0.09 \times 0.05765 \approx 0.2335$
Answer: $E [X] = 3$ ; $P (X = 2) \approx 0.233$ .

6. Common mistakes

Forgetting the binomial coefficient and only multiplying probabilities
Using $E [X] = n p$ correctly but then trying to re-derive it from the PMF under time pressure instead of stating it directly
Confusing Binomial (fixed $n$ trials) with Negative Binomial (fixed number of successes, random number of trials)
Assuming trials are independent when the problem describes sampling without replacement (that's Hypergeometric instead)

7. Quick check

If $X \sim Binomial (20, 0.5)$ , what is $Var (X)$ ?

Which scenario is best modeled by a Binomial distribution?

8. Practice

20 QuantSolve questions use this concept.

Practice this concept

9. Related methods

Symmetry Conditioning on the First Event

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