Negative Binomial Distribution

The negative binomial distribution is a generalization of the geometric distribution.
It describes the number of failures before the \( r^{th} \) success in a sequence of Bernoulli trials.
When \( r = 1 \), it reduces to the geometric distribution.

Definition of Random Variable

Let \( X \) denote the number of failures before the \( r^{th} \) success.
Each trial has:
- Probability of success = \( p \)
- Probability of failure = \( (1-p) \)
Then \( X = 0, 1, 2, \ldots \)
We are interested in \( P(X = x) \): the probability of observing \( x \) failures before the \( r^{th} \) success.

Derivation of Probability Mass Function (PMF)

To have \( x \) failures before the \( r^{th} \) success:
- The last (i.e., \( x + r \)-th) trial must be a success.
- The previous \( x + r – 1 \) trials must contain \( r – 1 \) successes and \( x \) failures.
Number of ways to arrange \( r – 1 \) successes in \( x + r – 1 \) trials:

\[
\binom{x + r – 1}{r – 1}
\]

Hence,

\[
P(X = x) = \binom{x + r – 1}{r – 1}(1-p)^x p^r
\]

Probability Mass Function (PMF)

\[
P(X = x) = \binom{x + r – 1}{r – 1} (1 – p)^x p^r, \quad x = 0, 1, 2, \ldots
\]

\( p \): probability of success
\( 1-p \): probability of failure
\( r \): number of required successes
\( \binom{x + r – 1}{r – 1} \): number of arrangements

Mean and Variance: Binomial Distribution

For the binomial distribution:
\[
\text{Mean} = np, \quad \text{Variance} = np(1-p)
\]
The probabilities for \( X = 0, 1, 2, \ldots \) are the successive terms of \( (q + p)^n \).

Mean and Variance: Negative Binomial (Mean)

The probabilities for \( X = 0, 1, 2, \ldots \) are successive terms of
\[
\left[ \frac{1}{p} + \left( \frac{q}{p} \right) \right]^r
\]
Thus, the mean is:
\[
\text{Mean} = \frac{rq}{p}
\]

Mean and Variance: Negative Binomial (Variance)

The variance is:
\[
\text{Var} = \frac{rq}{p^2}
\]

Example 1

Problem: Find the probability that the third head turns up in 5 tosses of an unbiased coin.

Given: \( p = \frac{1}{2}, \, r = 3, \, x + r = 5 \Rightarrow x = 2 \)

\[
P(X = x) = \binom{x + r – 1}{r – 1} p^r (1-p)^x
\]
\[
P(X = 2) = \binom{4}{2} \left( \frac{1}{2} \right)^3 \left( \frac{1}{2} \right)^2 = \frac{3}{16}
\]

Example 2

Problem: Find the probability that the third child in a family is the second daughter.

Given: \( p = \frac{1}{2}, \, r = 2, \, x + r = 3 \Rightarrow x = 1 \)

\[
P(X = 1) = \binom{2}{1} \left( \frac{1}{2} \right)^2 \left( \frac{1}{2} \right)^1 = \frac{1}{4}
\]

Example 3

A proof-reader catches a misprint with probability 0.8. Find the expected number of misprints in which he stops after catching the 20th misprint.

\[
E(X) = \frac{rq}{p} = \frac{20 \times 0.2}{0.8} = 5
\]
\[
E(X + r) = E(X) + r = 5 + 20 = 25
\]

Hence, the expected number of misprints is 25.

Example 4

Find the probability that at least 3 items are examined to get 2 defectives if \( p = 0.1, q = 0.9 \).

\[
P(X \ge 1) = 1 – P(X = 0)
\]
\[
P(X = 0) = {1 \choose 1} (0.1)^2 (0.9)^0 = 0.01
\]
\[
P(X \ge 1) = 1 – 0.01 = 0.99
\]

Hence, the probability is 0.99.

Example 5

A family stops after the second daughter. \( p = \frac{1}{2} \).

\[
E(X) = \frac{rq}{p} = \frac{2 \times \frac{1}{2}}{\frac{1}{2}} = 2
\]
\[
E(X + r) = E(X) + r = 2 + 2 = 4
\]

Expected total children = 4.

Applications

Modeling number of failures before a fixed number of successes.
Used in quality control and reliability testing.
Applied in ecology, insurance, and health studies.
Generalizes geometric distribution for multiple successes.

PDF

NegativeBinomial

Video

Connect with PostNetwork Academy

Website: www.postnetwork.co
YouTube: @postnetworkacademy
Facebook: facebook.com/postnetworkacademy
LinkedIn: linkedin.com/company/postnetworkacademy
GitHub: github.com/postnetworkacademy