Negative Binomial Distribution Table
The negative binomial distribution models the number of trials needed to achieve r successes, or the number of failures before the rth success.
Probability Formula
P(X = k) = C(k-1, r-1) × p^r × (1-p)^(k-r)
Where:
- k = total number of trials (k ≥ r)
- r = number of successes required
- p = probability of success on each trial
- C(n,k) = binomial coefficient
Probabilities for r = 2 Successes
P(X = k) - Trials needed for 2nd success
| k | p=0.10 | p=0.20 | p=0.25 | p=0.30 | p=0.40 | p=0.50 |
|---|---|---|---|---|---|---|
| 2 | 0.0100 | 0.0400 | 0.0625 | 0.0900 | 0.1600 | 0.2500 |
| 3 | 0.0180 | 0.0640 | 0.0938 | 0.1260 | 0.1920 | 0.2500 |
| 4 | 0.0243 | 0.0768 | 0.1055 | 0.1323 | 0.1728 | 0.1875 |
| 5 | 0.0292 | 0.0819 | 0.1055 | 0.1240 | 0.1382 | 0.1250 |
| 6 | 0.0328 | 0.0819 | 0.0989 | 0.1091 | 0.1037 | 0.0781 |
| 7 | 0.0354 | 0.0786 | 0.0892 | 0.0924 | 0.0746 | 0.0469 |
| 8 | 0.0372 | 0.0734 | 0.0781 | 0.0762 | 0.0522 | 0.0273 |
| 9 | 0.0383 | 0.0671 | 0.0668 | 0.0614 | 0.0358 | 0.0156 |
| 10 | 0.0388 | 0.0604 | 0.0563 | 0.0488 | 0.0241 | 0.0088 |
| 12 | 0.0383 | 0.0469 | 0.0387 | 0.0297 | 0.0106 | 0.0027 |
| 15 | 0.0349 | 0.0309 | 0.0212 | 0.0134 | 0.0030 | 0.0005 |
| 20 | 0.0275 | 0.0153 | 0.0081 | 0.0040 | 0.0005 | 0.0000 |
Probabilities for r = 3 Successes
P(X = k) - Trials needed for 3rd success
| k | p=0.10 | p=0.20 | p=0.25 | p=0.30 | p=0.40 | p=0.50 |
|---|---|---|---|---|---|---|
| 3 | 0.0010 | 0.0080 | 0.0156 | 0.0270 | 0.0640 | 0.1250 |
| 4 | 0.0027 | 0.0192 | 0.0352 | 0.0567 | 0.1152 | 0.1875 |
| 5 | 0.0049 | 0.0307 | 0.0527 | 0.0794 | 0.1382 | 0.1875 |
| 6 | 0.0073 | 0.0409 | 0.0659 | 0.0926 | 0.1382 | 0.1563 |
| 7 | 0.0098 | 0.0491 | 0.0742 | 0.0972 | 0.1244 | 0.1172 |
| 8 | 0.0122 | 0.0552 | 0.0781 | 0.0953 | 0.1037 | 0.0820 |
| 9 | 0.0145 | 0.0590 | 0.0781 | 0.0891 | 0.0815 | 0.0547 |
| 10 | 0.0163 | 0.0610 | 0.0756 | 0.0802 | 0.0611 | 0.0352 |
| 12 | 0.0190 | 0.0601 | 0.0663 | 0.0617 | 0.0322 | 0.0139 |
| 15 | 0.0201 | 0.0501 | 0.0477 | 0.0369 | 0.0118 | 0.0032 |
| 20 | 0.0182 | 0.0322 | 0.0243 | 0.0146 | 0.0023 | 0.0003 |
Probabilities for r = 5 Successes
P(X = k) - Trials needed for 5th success
| k | p=0.10 | p=0.20 | p=0.30 | p=0.40 | p=0.50 |
|---|---|---|---|---|---|
| 5 | 0.00001 | 0.0003 | 0.0024 | 0.0102 | 0.0313 |
| 6 | 0.00005 | 0.0012 | 0.0084 | 0.0307 | 0.0781 |
| 7 | 0.00013 | 0.0029 | 0.0177 | 0.0553 | 0.1094 |
| 8 | 0.00027 | 0.0058 | 0.0297 | 0.0774 | 0.1094 |
| 9 | 0.00048 | 0.0099 | 0.0424 | 0.0922 | 0.1094 |
| 10 | 0.00077 | 0.0150 | 0.0540 | 0.0983 | 0.0984 |
| 12 | 0.00163 | 0.0271 | 0.0700 | 0.0921 | 0.0688 |
| 15 | 0.00336 | 0.0432 | 0.0757 | 0.0648 | 0.0320 |
| 20 | 0.00571 | 0.0522 | 0.0584 | 0.0277 | 0.0074 |
| 25 | 0.00684 | 0.0456 | 0.0341 | 0.0088 | 0.0014 |
Cumulative Probabilities P(X ≤ k)
r = 2 Successes, P(X ≤ k)
| k | p=0.20 | p=0.30 | p=0.40 | p=0.50 |
|---|---|---|---|---|
| 2 | 0.0400 | 0.0900 | 0.1600 | 0.2500 |
| 3 | 0.1040 | 0.2160 | 0.3520 | 0.5000 |
| 4 | 0.1808 | 0.3483 | 0.5248 | 0.6875 |
| 5 | 0.2627 | 0.4723 | 0.6630 | 0.8125 |
| 6 | 0.3446 | 0.5814 | 0.7667 | 0.8906 |
| 7 | 0.4232 | 0.6738 | 0.8413 | 0.9375 |
| 8 | 0.4966 | 0.7500 | 0.8935 | 0.9648 |
| 10 | 0.6242 | 0.8507 | 0.9536 | 0.9893 |
| 15 | 0.8329 | 0.9647 | 0.9964 | 0.9998 |
| 20 | 0.9308 | 0.9924 | 0.9997 | 1.0000 |
Mean and Variance
| Parameter | Formula | Example (r=3, p=0.30) |
|---|---|---|
| Mean (μ) | r/p | 10 trials |
| Variance (σ²) | r(1-p)/p² | 23.33 |
| Standard Dev (σ) | √(r(1-p))/p | 4.83 |
Expected Number of Trials
E[X] = r/p
| r | p=0.20 | p=0.30 | p=0.40 | p=0.50 |
|---|---|---|---|---|
| 1 | 5.0 | 3.33 | 2.5 | 2.0 |
| 2 | 10.0 | 6.67 | 5.0 | 4.0 |
| 3 | 15.0 | 10.0 | 7.5 | 6.0 |
| 5 | 25.0 | 16.67 | 12.5 | 10.0 |
| 10 | 50.0 | 33.33 | 25.0 | 20.0 |
Example Problems
Example 1: Quality Control
Problem: A machine produces 20% defective items. What’s the probability that the 3rd defective item is found on the 8th inspection?
Solution (r = 3, p = 0.20, k = 8):
- P(X = 8) = 0.0552 (from table)
Example 2: Clinical Trials
Problem: If a treatment has 40% success rate, what’s the probability of achieving 5 successes within 10 patients?
Solution (r = 5, p = 0.40):
- P(X ≤ 10) = Sum of P(X=5) through P(X=10)
- ≈ 0.634
Special Cases
| When | Distribution becomes |
|---|---|
| r = 1 | Geometric distribution |
| p → 0, r → ∞, r×p → λ | Poisson distribution |
Relationship to Other Distributions
- Geometric: Special case when r = 1
- Binomial: Related through “successes in n trials” vs “trials for r successes”
- Poisson: Limit as p → 0
When to Use Negative Binomial
✓ Counting trials until rth success
✓ Quality control (defects found)
✓ Clinical trials (responses needed)
✓ Reliability (failures until r breakdowns)
✓ Overdispersed count data
Related Resources
- Negative Binomial Lesson - Complete guide
- Geometric Table - First success (r=1)
- Binomial Table - Fixed trials
- Poisson Table - Count data