Home / IB DP Maths / Application and Interpretation HL / IB Mathematics AI SL Binomial distribution. Mean and variance MAI Study Notes

IB Mathematics AI SL Binomial distribution. Mean and variance MAI Study Notes- New Syllabus

IB Mathematics AI SL Binomial distribution. Mean and variance MAI Study Notes

LEARNING OBJECTIVE

  • Binomial distribution.

Key Concepts: 

  • Binomial distribution
  • Mean and variance of the binomial distribution.

MAI HL and SL Notes – All topics

BINOMIAL DISTRIBUTION

◆ BINOMIAL DISTRIBUTION – \( B(n,p) \)

It is the distribution of a discrete random variable \( X \) which takes on the values:

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

with probability function:

$ p(x) = \binom{n}{x} p^x (1-p)^{n-x} \quad \text{for} \quad x = 0, 1, 2, 3, \dots, n $

where \( n \) and \( p \) are two parameters. We will see that the binomial distribution describes a certain type of problems.

SUCCESS with probability \( p \).
FAILURE with probability \( 1-p \).

 \( n \) = number of trials.
 \( p \) = probability of success.

While:

 \( X \) counts the number of (possible) successes.

\( X \sim B(n,p) \).

Since \( n \) is the number of trials, \( X \) can take on the values:

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

The probabilities \( P(X=0), P(X=1), P(X=2) \), etc., can be obtained by the GDC.

$ \mathbf{GDC}$

Our GDC (Casio) gives the results for a Binomial distribution:

$\text{MENU → Statistics → DIST → BINOMIAL:}$ We use Bpd or Bcd.

For simplicity, let us denote by:

Bpd(x): The probability of exactly \( x \) successes.
Bcd(x₁ to x₂): The probability from \( x₁ \) up to \( x₂ \) successes.

The menu for both functions is:

Data: Always Variable.
Numtrial: The number of trials, i.e., \( n \).
p: The probability of success \( p \) (for each game).

Then, for each value of \( x \) (or \( x₁ \) to \( x₂ \)), EXE gives the result.

Example 1

We toss a die 5 times. A success is defined as getting a six.

Given:

  • Number of trials: \( n = 5 \)
  • Probability of success: \( p = \frac{1}{6} \)

We may get 0, 1, 2, 3, 4, or 5 sixes.

Questions:

  1. What is the probability of getting 5 sixes in a row?
  2. What is the probability of getting no six at all?
  3. What is the probability of getting exactly 2 sixes and 3 no-sixes?
  4. What is the general formula to calculate the probability of getting exactly \( x \) sixes in 5 trials?
  5. What is the general binomial probability formula for any number of trials \( n \) and probability of success \( p \)?
▶️Answer/Explanation

Solution:

  1. Probability of getting 5 sixes in a row:
    $ \left( \frac{1}{6} \right)^5 $
  2. Probability of getting no six at all:
    $ \left( \frac{5}{6} \right)^5 $
  3. Probability of getting exactly 2 sixes and 3 no-sixes:
    $ \binom{5}{2} \left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^3 $
  4. General formula to find the probability of getting \( x \) sixes in 5 trials:
    $ \binom{5}{x} \left( \frac{1}{6} \right)^x \left( \frac{5}{6} \right)^{5-x} $
  5. General binomial probability formula for any \( n \) trials and probability \( p \):
    $ P(X = x) = \binom{n}{x} p^x (1 – p)^{n – x} $

MEAN AND VARIANCE OF THE BINOMIAL DISTRIBUTION

◆ EXPECTED VALUE/MEAN AND VARIANCE OF \( X \)

They are given by the formulae:

$ E(X) = np \quad \text{and} \quad \text{Var}(X) = np(1-p) $

For example (1)  above:

$ E(X) = 5 \times \frac{1}{6} = \frac{5}{6} \quad \text{and} \quad \text{Var}(X) = 5 \times \frac{1}{6} \times \frac{5}{6} = \frac{25}{36} $

EXAMPLE 2: Binomial Distribution Application

A box contains 5 balls: 1 BLACK and 4 WHITE. We win if we select a BLACK ball. We play this game 10 times.
Find:
(a) The probability to win exactly 4 times.
(b) The probability to win at most 4 times.
(c) The probability to win at least once.
(d) The expected number of winning games.
(e) The variance of the number of winning games.

▶️ Answer/Explanation

Solution:

The variable:
\( X = \text{number of winning games} \)
follows a binomial distribution with \( n=10 \) and \( p=\frac{1}{5}=0.2 \).
[We may also write \( X \sim B(10, 0.2) \).]

(a) The probability to win exactly 4 times is \( \text{Bpd}(4) = 0.088 \).

(b) The probability to win at most 4 times is \( \text{Bcd}(0 \text{ to } 4) = 0.967 \).
[In fact, \( P(X \leq 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) \).]

(c) The probability to win at least once is \( \text{Bcd}(1 \text{ to } 10) = 0.893 \).
[In fact, \( P(X \geq 1) = 1 – P(X=0) = 1 – 0.107 = 0.893 \).]

(d) The expected number is \( E(X) = np = 10 \times 0.2 = 2 \).

(e) The variance is \( \text{Var}(X) = np(1-p) = 10 \times 0.2 \times 0.8 = 1.6 \).

Scroll to Top