4.2.1 The Binomial Distribution

The binomial distribution describes a game with two outcomes, "win" or "lose." This game is to be played a fixed number of times irrespective of outcomes.

Notation

X \sim B(n, p)

The above notation describes such a game played n times where the probability of success on each go is $p$ . $X$ refers to the number of wins obtained.

X \sim B(n, p)

$p$ : This is the probability of a win each time.
$n$ : Number of trials.
$X$ : Number of wins. (Has a binomial distribution)

Assumptions

Trials are independent.
Fixed probability of winning each time.

infoNote

Example: Consider a game in which a dice is rolled $20$ $times$ . Let $X$ be the number of "sixes" rolled in total.

Consider $X$ to be the number of "sixes".
We have $P(\text{win}) = \frac{1}{6}$ .

Thus,

X \sim B(20, \frac{1}{6})

In the previous scenario, find the probability we roll exactly seven 6s in the twenty throws.

P(X = 7) = \left(\frac{1}{6}\right)^7 \left(\frac{5}{6}\right)^{13} \times \binom{20}{7} = :highlight[0.02588]

Explanation:

Seven $6s$
Seven wins
Thirteen losses
Number of different orders events can happen in

The Binomial Distribution

\text{If}\ X \sim B(n, p)\ \text{then}\ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}, \\ \text{mean of}\ X \text{is}\ :highlight[np],\text{variance of}\ X\ \text{is}\ :highlight[np(1-p)].

\text{Where}\ ^n{C_r} = _nC_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}

infoNote

Example: The random variable $X \sim B(8, \frac{1}{3})$ .

Questions:

Find:

a) $P(X = 2)$

b) $P(X = 5)$

c) $P(X \leq 1)$

Find $P(X = 2)$ Step 1: Use the Binomial Probability Formula

For a binomial distribution, the probability of getting r successes in n trials is given by:

P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}

Where:

$n = 8$ (the number of trials)
$p = \frac{1}{3}$ (the probability of success on each trial)
$r = 2$ (the number of successes we want)

Step 2: Substitute the Values and Calculate the Result

P(X = 2) = \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^6 \times C_2^8

P(X = 2)= \frac{792}{6561} \approx :success[0.273]

Instructions for Calculator:

Go to " $Dist$ " by pressing [ $SHIFT$ ] then [ $1$ ].
Select [ $4:Binomial$ $PD$ ].
Enter $X = 2, n = 8, p = 1/3$ .
Calculate $P = :success[0.2731290962]$ .

b. Find $P(X=5)$

P(X = 5) = \left(\frac{1}{3}\right)^5 \left(\frac{2}{3}\right)^3 \times C_5^8

P(X=5)= \frac{468}{6561} \approx :success[0.06828]

c. Find $P(X \leq 1)$

Step 1: Understand the Question

We are asked to find the probability that X is less than or equal to 1.

This can be shown in this notation:

P(X \leq 1) = P(X = 0) + P(X = 1)

Step 2: Use the Binomial Formula for $P(X=0)$ and $P(X=1)$

For $P(X=0)$

P(X = 0) = \left(\frac{1}{3}\right)^0 \left(\frac{2}{3}\right)^8 \times ^8{C_0}

For $P(X=1)$

P(X=1)=\left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^7 \times ^8{C_1}

Step 3: Calculate the Result

P(X \leq 1) = \frac{1280}{6561} \approx :success[0.1951]

infoNote

Example: The probability of a switch being faulty is $0.08$ . A random sample of $10$ switches is taken from the production line.

Questions:

a) Define a suitable distribution to model the number of faulty switches in this sample, and justify your choice.

b) Find the probability that the sample contains $4$ faulty switches.

a) Define a suitable distribution to model the number of faulty switches in this sample, and justify your choice.

Step 1: Identify the Distribution

We are asked to model the number of faulty switches in a random sample of $10$ switches, with the probability of any switch being faulty given as $0.08$ .

This is a binomial distribution because:

There is a fixed number of trials ( $10$ switches)
Each switch is either faulty or not faulty (a binary outcome)
The probability of a switch being faulty is constant at $0.08$
The switches are independent of each other. Thus, the number of faulty switches X follows the binomial distribution:

X \sim B(10, 0.08)

Where:

$10$ is the number of trials (switches)
$0.08$ is the probability of success (faulty switch)

Step 2: Justify the Choice

The trials (checking each switch) are independent because whether one switch is faulty does not affect the others.
The probability of finding a faulty switch remains constant at $0.08$ for each switch.

b) Find the probability that the sample contains 4 faulty switches.

Step 1: Use the Binomial Formula

We need to calculate $P(X=4)$ where $X \sim B(10, 0.08)$

P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}

Where:

$n = 10$ (the number of switches)
$p = 0.08$ (the probability of a switch being faulty)
$r = 4$ (the number of faulty switches)

Step 2: Substitute the Values and Calculate the Result

P(X = 4) = 0.08^4 \times 0.92^6 \times ^{10}{C}_4

= 210 \times (0.08)^4 \times (0.92)^6

= :success[5.216 \times 10^{-3}]

Thus, the probability that the sample contains $4$ faulty switches is approximately $0.005216$ or $5.216 \times 10^{-3}$

The Binomial Distribution Simplified Revision Notes for A-Level AQA Maths Statistics

4.2.1 The Binomial Distribution

Notation

Assumptions

The Binomial Distribution

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