Which of these is a Bernoulli trial?

Tossing a coin and recording heads or not

Rolling a die and recording the number

Which situation is NOT binomial?

Draw 5 cards without replacement, count hearts

20 free throws with a steady 60% success rate, count goals

10 seeds each with an 80% chance to sprout, count sprouts

The Binomial Distribution | HSC Maths Extension 1

In short: The binomial distribution counts successes in $n$ independent trials that each have the same probability $p$ of success. $P (X = r) = (r n) p^{r} (1 - p)^{n - r}$ , with mean $n p$ and variance $n p (1 - p)$ .

Starting point: what you already know

Probability basics: $P (not A) = 1 - P (A)$ , and for independent events $P (A and B) = P (A) \times P (B)$ .
Combinations (Combinations chapter): $(r n) = \frac{n !}{r ! ( n - r )!}$ counts the ways to choose $r$ things from $n$ .
Discrete random variables (Mathematics Advanced): a table of values and probabilities that add to 1, with mean $E (X) = \sum x P (X = x)$ and variance $Var (X) = E (X^{2}) - [E (X)]^{2}$ .
The binomial theorem: $(q + p)^{n} = \sum (r n) p^{r} q^{n - r}$ .

Bernoulli trials

A Bernoulli trial has exactly two outcomes, called success and failure. The probability of success is $p$ , and of failure is $1 - p$ , sometimes written $q$ .

Let $X = 1$ for a success and $X = 0$ for a failure. Then:

$E (X) = 0 \times (1 - p) + 1 \times p = p$
$Var (X) = E (X^{2}) - p^{2} = p - p^{2} = p (1 - p)$

The binomial distribution

Now repeat a Bernoulli trial $n$ times and count the successes, $X$ . This is a binomial distribution, written $X \sim Bin (n, p)$ , when:

there are a fixed number $n$ of trials,
each has only two outcomes,
the trials are independent,
$p$ is the same every time.

Why it works: one particular order of $r$ successes and $n - r$ failures has probability $p^{r} (1 - p)^{n - r}$ . There are $(r n)$ different orders.

Bin(10, 0.3). The most likely number of successes is 3, which is also the mean, 10 × 0.3.

Mean and variance

This makes sense: $X$ is the sum of $n$ separate Bernoulli trials, each with mean $p$ and variance $p (1 - p)$ .

For the basketball player: $E (X) = 3$ shots, $Var (X) = 2.1$ , $σ \approx 1.45$ shots.

Binomial distribution

Bernoulli trial

Two outcomes
Mean p
Variance p(1 - p)

Conditions

Fixed n
Two outcomes
Independent
Same p

Formula

nCr p^r (1 - p)^(n - r)
nCr counts the orders

Summary numbers

Mean np
Variance np(1 - p)

Tricks

At least one: 1 - P(none)
Use complements

Year 12Working backwards from the mean and variance

$X \sim Bin (n, p)$ has mean 6 and variance 4.2. Find $n$ and $p$ .

Divide: $\frac{n p ( 1 - p )}{n p} = \frac{4.2}{6}$ , so $1 - p = 0.7$ and $p = 0.3$ . Then $n = \frac{6}{0.3} = 20$ .

More facts from the mean and variance. $E (X^{2}) = Var (X) + [E (X)]^{2}$ , so for $X \sim Bin (10, 0.5)$ , $E (X^{2}) = 2.5 + 25 = 27.5$ . The variance of one Bernoulli trial, $p (1 - p)$ , is largest at $p = 0.5$ , where it is 0.25.

The most likely value. Compare neighbouring probabilities: $\frac{P ( X = k + 1 )}{P ( X = k )} = \frac{n - k}{k + 1} \cdot \frac{p}{1 - p}$ . The probabilities rise while this ratio is above 1. For $Bin (10, 0.3)$ the most likely value is 3, close to the mean $n p$ .

Two or more. $P (X \geq 2) = 1 - P (X = 0) - P (X = 1)$ .

A normal approximation. For large $n$ , $Bin (n, p)$ is close to a normal distribution with mean $n p$ and standard deviation $n p (1 - p)$ . For $Bin (100, 0.5)$ , the mean is 50 and the standard deviation 5, so getting 60 or more ( $z = 2$ ) has a chance of about 2.5%.

Quick summary

A Bernoulli trial has two outcomes, with mean $p$ and variance $p (1 - p)$ .
Binomial needs: fixed $n$ , two outcomes, independent trials, same $p$ .
$P (X = r) = (r n) p^{r} (1 - p)^{n - r}$ .
$E (X) = n p$ and $Var (X) = n p (1 - p)$ .
"At least one" is $1 - P (X = 0)$ .

Quiz

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Try 3 questions now

Which of these is a Bernoulli trial?
1. A Rolling a die and recording the number
2. B Tossing a coin and recording heads or not
3. C Measuring a person's height
4. D Counting cars in an hour
Show answer
B: Tossing a coin and recording heads or not
It has exactly two outcomes.
A Bernoulli variable has $p = 0.4$ . Its variance is:
1. A 0.4
2. B 0.24
3. C 0.16
4. D 0.6
Show answer
B: 0.24
$p (1 - p) = 0.4 \times 0.6$ .
Which situation is NOT binomial?
1. A 8 coin tosses, count heads
2. B 20 free throws with a steady 60% success rate, count goals
3. C Draw 5 cards without replacement, count hearts
4. D 10 seeds each with an 80% chance to sprout, count sprouts
Show answer
C: Draw 5 cards without replacement, count hearts
Without replacement, $p$ changes after each draw.

The Binomial Distribution