---
title: "Negative binomial: 5th hit on 10th throw"
subject: "Probability and Statistics"
chapter: 2
key: "semester_5/probability_and_statistics:nb-hit-10th-throw-5th-hit"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/nb-hit-10th-throw-5th-hit
site: BCT Engineering
---

# Negative binomial: 5th hit on 10th throw

Probability and Statistics · Chapter 2: Discrete Probability Distribution

- Asked in: 2073 Shrawan, 5 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/nb-hit-10th-throw-5th-hit
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-2

## Question

Define negative binomial Distribution. If a boy is throwing stone at a target what is the probability that his 10th throw is his 5th hit, if the probability of hitting the target at any trial is 0.6. Also find the mean and variance of random variable.

## Short answer

Negative binomial distribution gives the trial number of the k-th success. For 10th throw as 5th hit with p=0.6, probability=0.1003, mean=8.33, variance=5.56.

## Answer

#### Negative binomial distribution
In independent Bernoulli trials with constant success probability p, let X be the trial number on which the k-th success occurs. Then X has a **negative binomial distribution**:

P(X = x) = C(x − 1, k − 1) pᵏ qˣ⁻ᵏ, x = k, k + 1, k + 2, …

Mean = k/p, variance = kq/p². (For k = 1 it reduces to the geometric distribution.)

#### Probability that the 10th throw is the 5th hit
Here k = 5, x = 10, p = 0.6, q = 0.4. The 10th throw must be a hit, and exactly 4 hits must occur in the first 9 throws.

P(X = 10) = C(9, 4) (0.6)⁵ (0.4)⁵
= 126 × 0.07776 × 0.01024
= **0.1003**

#### Mean and variance
Mean = k/p = 5/0.6 = **8.33 throws**

Variance = kq/p² = 5 × 0.4 / 0.36 = 2/0.36 = **5.56**

**Answer: probability = 0.1003, mean = 8.33, variance = 5.56**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. X in a negative binomial distribution is:
   A. The number of successes in k trials
   B. The number of failures before any success
   C. The trial number of the k-th success
   D. A rare-event count per interval
   Answer: C. It counts trials until the k-th success. (Key point: Definition)

2. For the 10th throw to be the 5th hit we need:
   A. 5 hits in the first 9 throws
   B. 4 hits in the first 10 throws
   C. A hit on throw 10 and exactly 4 hits in the first 9
   D. A miss on throw 10
   Answer: C. The last trial is the k-th success; k - 1 successes occur earlier, giving C(9,4). (Key point: Setup)

3. P(X = 10) is:
   A. C(9,4)(0.6)^5(0.4)^5 = 0.1003
   B. C(10,5)(0.6)^5(0.4)^5 = 0.2007
   C. C(9,5)(0.6)^5(0.4)^4
   D. C(9,4)(0.6)^4(0.4)^5
   Answer: A. Use C(x-1,k-1) p^k q^(x-k) with x = 10, k = 5. (Key point: Probability)

4. The mean and variance are:
   A. kp = 3 and kpq = 1.2
   B. 1/p = 1.67 and q/p^2 = 1.11
   C. k/q = 12.5 and kp/q^2
   D. k/p = 8.33 and kq/p^2 = 5.56
   Answer: D. Mean = 5/0.6 = 8.33; variance = 5 x 0.4 / 0.36 = 5.56. (Key point: Mean and variance)

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