4.10 Binomial Distribution

1. Description of the Problem

The Binomial distribution describes the probability of a discrete random variable $X$. It applies to experiments with exactly two possible outcomes:

  • SUCCESS (with constant probability $p$)
  • FAILURE (with probability $1-p$)

We repeat the experiment $n$ independent times. The parameters are:

$n$ = total number of trials
$p$ = probability of success in a single trial

The variable $X$ counts the number of successes. We say $X$ follows a binomial distribution, written $X \sim B(n,p)$. Since $n$ is the number of trials, $X$ takes the values:

$0, 1, 2, 3, 4, \dots, n$

EXAMPLE 1

We toss a die 5 times. Success is rolling a six. Here, $n=5$ and $p=\dfrac{1}{6}$. We can have 0, 1, 2, 3, 4, or 5 successes.

$X$ 0 1 2 3 4 5
$P(X=x)$ 0.4019 0.4019 0.1608 0.0322 0.0032 0.0001

We can answer inequality questions using cumulative probabilities:

Find the probability of: Notation Result
exactly 3 sixes $P(X=3)$ 0.0322
at most 3 sixes $P(X \le 3)$ 0.9967
less than 3 sixes $P(X < 3)$ 0.9645
more than 3 sixes $P(X > 3)$ 0.0033
at least 3 sixes $P(X \ge 3)$ 0.0355

2. The Binomial Formula

Remark: This formula helps conceptualize binomial probabilities.

  • The probability of obtaining 5 sixes in a row is: $\left(\dfrac{1}{6}\right)^5$
  • The probability of obtaining 0 sixes is: $\left(\dfrac{5}{6}\right)^5$
  • The probability of obtaining exactly 2 sixes and 3 no-sixes is: $^5C_2 \left(\dfrac{1}{6}\right)^2 \left(\dfrac{5}{6}\right)^3$

Here, $^5C_2$ is the number of ways to arrange 2 sixes in 5 trials. In general, for $n$ trials with a success probability of $p$, $P(X=x)$ is:

$P(X=x) = {^n}C_x p^x (1-p)^{n-x}$

3. Expected Value and Variance

For a random variable $X \sim B(n, p)$, the expected value (mean) and variance are:

$E(X) = np$
$Var(X) = np(1-p)$
For the dice example ($n=5$, $p=\dfrac{1}{6}$):
$E(X) = 5\left(\dfrac{1}{6}\right) = \mathbf{\dfrac{5}{6}}$
$Var(X) = 5\left(\dfrac{1}{6}\right)\left(\dfrac{5}{6}\right) = \mathbf{\dfrac{25}{36}}$
Notice (HL Only): If you know $E(X)$ and $Var(X)$, you can find $E(X^2)$:
$$ \begin{aligned} Var(X) &= E(X^2) - (E(X))^2 \\ E(X^2) &= Var(X) + (E(X))^2 \\ E(X^2) &= np(1-p) + (np)^2 \end{aligned} $$

EXAMPLE 2

A box contains 5 balls: 1 BLACK and 4 WHITE. We win if we select a BLACK ball. We play this game 10 times (with replacement). Find the probabilities.

The variable $X = \text{"number of winning games"}$ follows a binomial distribution with $n=10$ and $p=\dfrac{1}{5}=0.2$. We write $X \sim B(10, 0.2)$.
(a) Probability of winning exactly 4 times:
This is $\mathbf{0.088}$.
[$P(X=4) = {^{10}C_4}(0.2)^4(0.8)^6 = 0.088$]
(b) Probability of winning at most 4 times:
This cumulative probability is $\mathbf{0.967}$.
[$P(X \le 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)$]
(c) Probability of winning at least once:
This is $\mathbf{0.893}$.
[Alternatively: $P(X \ge 1) = 1 - P(X=0) = 1 - 0.107 = 0.893$]
(d) Expected number of winning games:
$E(X) = np = 10 \times 0.2 = \mathbf{2}$
(e) Variance of winning games:
$Var(X) = np(1-p) = 10 \times 0.2 \times 0.8 = \mathbf{1.6}$

EXAMPLE 3

Let $p=0.2$. Given $P(X=1) = 0.268$, find $n$.

Solution: $n$ must be a positive integer.
Testing integer values shows that $n=10$ results in 0.268.
Hence, $\mathbf{n=10}$.

4. Mode of a Binomial DistributionHL ONLY

To find the mode (the outcome with the highest probability), first calculate $E(X) = np$. The mode is adjacent to the mean.

If the expected value is a decimal:
Say $n=20$ and $p=\dfrac{1}{6}$, meaning $E(X) = \dfrac{20}{6} \approx 3.33$.
Evaluate the probabilities for the nearest integers (3 and 4):
$P(X=3) = 0.237$
$P(X=4) = 0.202$
Since $0.237 > 0.202$, the highest probability is at $X=3$. The mode is 3.
If the expected value is a whole number:
Say $n=60$ and $p=\dfrac{1}{6}$, meaning $E(X) = \dfrac{60}{6} = 10$.
Evaluate the probabilities for adjacent integers (9, 10, 11):
$P(X=9) = 0.134$
$P(X=10) = 0.137$
$P(X=11) = 0.126$
The peak probability is at the mean. The mode is 10.
Notice (Two Modes Case): Sometimes there are two modes.
For $n=5$ and $p=\dfrac{1}{6}$, $E(X) = \dfrac{5}{6} \approx 0.833$. Check the adjacent integers 0 and 1:
$P(X=0) = 0.4019$
$P(X=1) = 0.4019$
The probabilities are identical. Therefore, there are two modes: 0 and 1.