The Binomial Distribution

The Binomial Distribution — Moosa Academy

A binomial experiment repeats the same trial a fixed number of times, where each trial has only two outcomes — success or failure. The binomial distribution then gives the probability of getting exactly $x$ successes.

Conditions When is it binomial?

Two outcomes: each trial ends in success or failure.
Fixed $n$ : the number of trials is known in advance.
Constant $p$ : the success probability is the same on every trial.
Independence: the result of one trial does not affect the others.

Typical examples: tossing a coin $n$ times $(p = 0.5)$ ; inspecting $n$ items on a production line $(p = 0.95)$ ; answering $n$ true/false questions $(p = 0.7)$ .

Formula The binomial probability

The probability of exactly $x$ successes in $n$ trials is:

P(X = x) = \binom{n}{x}\, p^{x}\, q^{\,n-x}

n

= number of trials,

x

= number of successes wanted.

p

= probability of success,

q = 1 - p

= probability of failure.

\binom{n}{x}

counts the ways to place

x

successes among

n

trials.

Example A coin tossed 6 times

Exactly 3 heads, with $n = 6$ , $x = 3$ , $p = 0.5$ :

\binom{6}{3}(0.5)^3(0.5)^3 = \dfrac{20}{64}

⟹ P(X = 3) = 31.25%.

Example Two more

A die rolled 10 times — exactly two sixes

Here

n = 10

x = 2

p = \tfrac{1}{6}

\binom{10}{2}\left(\tfrac{1}{6}\right)^{2}\left(\tfrac{5}{6}\right)^{8}

⟹ P(X = 2) ≈ 29.1%.

A production line (5% defective) — inspect 20, exactly one defective

Here

n = 20

x = 1

p = 0.05

\binom{20}{1}(0.05)^{1}(0.95)^{19}

⟹ P(X = 1) ≈ 37.7%.

Note Expected number of successes

On average, a binomial experiment gives $\mu = n \times p$ successes. For the coin above, $\mu = 6 \times 0.5 = 3$ — which is exactly where the distribution peaks.

Reference The three factors

Part	Symbol	Meaning
Combinations	C(n, x)	ways to place x successes in n trials
Success probability	pˣ	x successes occurring
Failure probability	qⁿ⁻ˣ	(n − x) failures occurring
Expected value	μ = np	average number of successes

Summary

Conditions: two outcomes, a fixed $n$ , a constant $p$ , and independent trials.
The law: $P(X = x) = \binom{n}{x}\,p^{x}\,q^{\,n-x}$ — three factors multiplied together.
$\binom{n}{x}$ is the number of ways to spread $x$ successes across $n$ trials.
Expected value: $\mu = np$ — a larger $p$ or $n$ raises the average number of successes.