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105 LearnersLast updated on October 22, 2025

The negative binomial distribution tells us how many trials it takes until we reach a certain number of successes, focusing on the trial in which the final success occurs. In this article, we’ll explore this concept in more detail.
The negative binomial distribution models the number of trials needed to achieve r successes, assuming each trial has the same probability of success, θ. Here, successes are fixed and trials are variable. If you repeat independent trials until the rᵗʰ success occurs, the probability that this happens after x failures (i.e., on the x + rth trial) is given by:
\(f(x) = \left( \frac{x + r - 1}{r - 1} \right) p r^x, \quad x = 0, 1, 2, \dots \)
Where:
x = number of failures before the r-th success
r = total number of successes we want to observe
p = probability of success on a single trial
q = probability of failure on a single trial = 1 − p
The objective of the negative binomial distribution is to help us figure out how many trials it might take to get a certain number of successes when each trial has the same chance of success. It depends mainly on two things:
For example:
Let’s say a basketball player wants to make 4 successful free throws.
This setup tells us how many total attempts we might expect before the player gets those 4 makes.
The properties of the negative binomial distribution are as follows:
The probability density function of the negative binomial distribution gives the probability of observing a specific number of failures before reaching a fixed number of successes. It is given by:
\(P(X = x) = \left( \frac{x + r - 1}{r - 1} \right) p^r (1 - p)^x, \quad x = 0, 1, 2, \dots \)
Where,
x = number of failures before rth success
r = required number of successes
p = probability of success in each trial
(1-p) = q = probability of failure, and
The binomial coefficient shows how many different ways the failures and successes can be ordered before the final success occurs.
It describes the probability that the rth success occurs exactly on trial x + r.
The mean and variance tell us what to expect from the number of trials needed to reach a set number of successes.
Example:
If a player has a 40% chance of scoring on each attempt and wants to score 3 times, then:
While working with the negative binomial distribution, students may confuse it with the binomial distribution, among other common mistakes. In this section, we will identify these mistakes and learn how to avoid repeating them.
Negative binomial distribution has a wide range of applications in various fields like hospitality, R&D, and marketing. Here, we’ve mentioned a few applications.
A basketball player makes a successful shot with a probability of 0.6. What is the probability that she makes her 3rd successful shot on the 5th attempt?
0.20736
We will use the negative binomial distribution, which gives the probability that the r-th success happens on the x-th trial.
Here:
r = 3 (we want the 3rd success)
x = 5 (shot is successful on the 5th attempt)
p = 0.6 (probability of success)
q = 1 - p = 0.4 (probability of failure)
The negative binomial probability formula is:
\(P(X = x) = \left( \frac{x - 1}{r - 1} \right) p^r q^{x - r} \)
Substituting the values of x, r, p, and q in the formula, we get:
\(P(X = 5) = \binom{4}{2} (0.6)^3 (0.4)^2 = 6 \cdot 0.216 \cdot 0.16 = 0.20736 \)
A fair coin (p = 0.5) is tossed until we get 3 heads. What is the probability that this takes exactly 5 tosses?
0.1875
We’re tossing a fair coin and want the third head to happen on the 5th toss. That means:
So, we’re counting the number of failures (tails) before we reach the 3rd success (head).
\(P(X = 2) = \left( \frac{2 + 3 - 1}{3 - 1} \right) (0.5)^3 (0.5)^2 = \binom{4}{2} \cdot 0.125 \cdot 0.25 = 6 \cdot 0.03125 = 0.1875 \)
A machine produces 10% defective items. What is the probability that the 2nd non-defective (success) item is found on the 5th trial?
0.00324
We’re looking for the probability that the 2nd non-defective item comes up on the 5th trial. That means:
The first 4 items must include exactly 1 non-defective and 3 defective items
The 5th item must be non-defective
We’re using the negative binomial distribution, where:
r = 2
p = 0.9
x = 3
Substituting the values in the formula, we get:
\(P(X = 3) = \left( \frac{3 + 2 - 1}{2 - 1} \right) (0.1)^3 (0.9)^2 = \binom{4}{1} \cdot 0.001 \cdot 0.81 = 4 \cdot 0.00081 = 0.00324
\)
A customer makes a purchase 40% of the time. What’s the probability that the 3rd purchase happens on the 7th visit?
15⋅0.0082944 ≈ 0.1244
p = 0.4, r = 3, total trials = 7 ⇒ x = 4
\(P(X = 4) = \left( \frac{4 + 3 - 1}{2} \right) \cdot (0.6)^4 \cdot (0.4)^3 = \binom{6}{2} \cdot 0.1296 \cdot 0.064
\)
In a drug trial, each test has a 20% chance of success. What is the probability that the 5th success occurs on the 12th trial?
≈0.0221
We want the probability that the 5th success happens exactly on the 12th trial. That means:
In the first 11 trials, there must be 4 successes and 7 failures (in any order)
The 12th trial must be a success
To solve this, we use the following formula:
\(P(X = 7) = \left( \frac{7 + 5 - 1}{4} \right) \cdot (0.8)^7 \cdot (0.2)^5 = \binom{11}{4} \cdot 0.2097152 \cdot 0.00032 \)
so: 330 ⋅ 0.2097152 ⋅ 0.00032 ≈ 0.0221
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.






