Geometric & Negative Binomial Distributions Revision Notes for Edexcel A-Level Further Mathematics

17.1.4 The Negative Binomial Distribution

Introduction

The negative binomial distribution is a probability distribution that models the number of trials needed to achieve a fixed number of successes ( $r$ ) in a sequence of independent and identically distributed Bernoulli trials, each with success probability $p$ .

It is often used to model scenarios where you are counting failures before achieving a specified number of successes. This note covers:

The probability mass function (PMF).
Relationships with the binomial distribution.
Mean and variance of the negative binomial distribution.
Applications and examples.

Probability Mass Function (PMF)

If $Y \sim \text{NegBin}(r, p)$ , the probability that $Y = x$ (i.e., $x$ failures before achieving $r$ successes) is given by:

P(Y = x) = \binom{x + r - 1}{r - 1} p^r (1-p)^x

where:

$r$ is the fixed number of successes,
$p$ is the probability of success in a single trial,
$\binom{x + r - 1}{r - 1} = \frac{(x + r - 1)!}{(r - 1)! \, x!}$ is the number of ways to arrange $r-1$ successes and $x$ failures.

Relationship with the Binomial Distribution

There is an important relationship between the negative binomial distribution and the binomial distribution:

If $Y \sim \text{NegBin}(r, p)$ , the cumulative probability $P(Y \leq n)$ can be related to a binomial random variable $X \sim \text{Bin}(n, p)$

P(Y \leq n) = P(X \geq r)

where $X$ is the number of successes in $n$ trials.

This relationship helps in solving cumulative probability problems.

Mean and Variance

For $Y \sim \text{NegBin}(r, p)$

Mean ( $\mu$ )

\mu = \frac{r(1-p)}{p}

Variance ( $\sigma^2$ )

\sigma^2 = \frac{r(1-p)}{p^2}

Worked Examples

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Example 1: Finding Probabilities with the Negative Binomial Distribution

Problem

A basketball player makes a shot with a success probability of $p = 0.6$ .

What is the probability that it takes exactly 4 shots to make 2 successful shots?

Solution

The number of failures before achieving $r = 2$ successes follows a negative binomial distribution:

P(Y = x) = \binom{x + r - 1}{r - 1} p^r (1-p)^x

Step 1**:** Identify values:

$r = 2$ (number of successes),
$p = 0.6$ (probability of success),
$x = 2$ (number of failures before achieving 2 successes)

Step 2**:** Use the PMF formula:

P(Y = 2) = \binom{2 + 2 - 1}{2 - 1} (0.6)^2 (1-0.6)^2

Step 3**:** Calculate:

P(Y = 2) = \binom{3}{1} (0.6)^2 (0.4)^2

P(Y = 2) = 3 \times 0.36 \times 0.16 = 0.1728

Final Answer:

The probability that it takes exactly $4$ shots ( $2$ failures) to make $2$ successful shots is 0.1728.

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Example 2: Using the Relationship Between Negative Binomial and Binomial Distributions

Problem

A manufacturing process produces defective items with a probability of $p = 0.2$

Find the probability that it takes at most 7 trials to produce 3 non-defective items.

Solution

This is a negative binomial problem with $r = 3$ successes and $p=0.8$ (probability of non-defective items). The probability $P(Y \leq 7)$ can be written using the relationship with the binomial distribution:

P(Y \leq 7) = P(X \geq 3)

where $X \sim \text{Bin}(7, 0.8)$ is the number of non-defective items in 7 trials.

Step 1: Use the complement rule:

P(X \geq 3) = 1 - P(X \leq 2)

Step 2: Find $P(X \leq 2)$ using the binomial formula:

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

For $n=7,p=0.8$

P(X = 0) = \binom{7}{0} (0.8)^0 (0.2)^7 = 1 \times 0.2^7 = 0.000128

P(X = 1) = \binom{7}{1} (0.8)^1 (0.2)^6 = 7 \times 0.8 \times 0.2^6 = 0.005376

P(X = 2) = \binom{7}{2} (0.8)^2 (0.2)^5 = 21 \cdot 0.64 \cdot 0.2^5 = 0.057344

Step 3: Sum probabilities:

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

P(X \leq 2) = 0.000128 + 0.005376 + 0.057344 = 0.062848

Step 4: Find $P(X \geq 3)$

P(X \geq 3) = 1 - P(X \leq 2) = 1 - 0.062848 = 0.937152

Final Answer:

The probability that it takes at most $7$ trials to produce $3$ non-defective items is 0.9372

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Example 3: Mean and Variance of a Negative Binomial Distribution

Problem

A factory produces defective items with a probability of $p = 0.3$ . If we are interested in finding the number of trials needed to produce 5 defective items, find the mean and variance of the distribution.

Solution

For a negative binomial distribution:

\mu = \frac{r(1-p)}{p}, \quad \sigma^2 = \frac{r(1-p)}{p^2}

Step 1: Mean:

\mu = \frac{r(1-p)}{p} = \frac{5(1-0.3)}{0.3} = \frac{5 \times 0.7}{0.3} = \frac{3.5}{0.3} = 11.67

Step 2: Variance:

\sigma^2 = \frac{r(1-p)}{p^2} = \frac{5(1-0.3)}{(0.3)^2} = \frac{5 \times 0.7}{0.09} = \frac{3.5}{0.09} \approx 38.89

Final Answer:

Mean: 11.67
Variance: 38.89

Note Summary

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Common Mistakes

Misinterpreting $x$ and $r$ : $x$ is the number of failures, while $r$ is the number of successes.
Confusing binomial and negative binomial distributions: Negative binomial counts failures before successes, while binomial counts successes in a fixed number of trials.
Forgetting to adjust for cumulative probabilities: Use $P(Y \leq n) = P(X \geq r)$ when needed.
Incorrectly applying mean and variance formulas: Ensure $p$ and $r$ are correctly substituted.

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Key Formulas

PMF:

P(Y = x) = \binom{x + r - 1}{r - 1} p^r (1-p)^x

Mean:

\mu = \frac{r(1-p)}{p}

Variance:

\sigma^2 = \frac{r(1-p)}{p^2}

Relationship with Binomial:

P(Y \leq n) = P(X \geq r)

Geometric & Negative Binomial Distributions (Edexcel A-Level Further Mathematics): Revision Notes

17.1.4 The Negative Binomial Distribution

Introduction

Probability Mass Function (PMF)

Relationship with the Binomial Distribution

Mean and Variance

Mean ( $\mu$ )

Variance ( $\sigma^2$ )

Worked Examples

Example 1: Finding Probabilities with the Negative Binomial Distribution

Example 2: Using the Relationship Between Negative Binomial and Binomial Distributions

Example 3: Mean and Variance of a Negative Binomial Distribution

Note Summary

Common Mistakes

Key Formulas

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