2. Consider a Geometric Distribution with probability of success of 0.0025 D. Compute P(10 trial is a success)
2. Consider a Geometric Distribution with probability of success of 0.0025 D. Compute P(10 trial is a success)
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![**Geometric Distribution Problem**
**2. Consider a Geometric Distribution with a probability of success of 0.0025.**
**D. Compute P(10th trial is a success).**
**Explanation:**
A geometric distribution models the number of trials needed to get the first success in repeated, independent Bernoulli trials, each with the same probability of success. In this problem, you're asked to find the probability that the first success occurs on the 10th trial when the probability of success in each individual trial is 0.0025.
To solve this, use the formula for the probability that the first success occurs on the n-th trial:
\[ P(X = n) = (1 - p)^{n-1} \cdot p \]
where:
- \( n \) is the trial number,
- \( p \) is the probability of success on a single trial,
- \( 1 - p \) is the probability of failure on a single trial.
Substitute \( p = 0.0025 \) and \( n = 10 \) into the formula to compute the desired probability.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F69da6d7b-7352-4dea-8ba3-a40ad4f9d8c3%2Fc634a435-5fde-4f98-bb34-f0d58e96e68f%2Fexjenpf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Geometric Distribution Problem**
**2. Consider a Geometric Distribution with a probability of success of 0.0025.**
**D. Compute P(10th trial is a success).**
**Explanation:**
A geometric distribution models the number of trials needed to get the first success in repeated, independent Bernoulli trials, each with the same probability of success. In this problem, you're asked to find the probability that the first success occurs on the 10th trial when the probability of success in each individual trial is 0.0025.
To solve this, use the formula for the probability that the first success occurs on the n-th trial:
\[ P(X = n) = (1 - p)^{n-1} \cdot p \]
where:
- \( n \) is the trial number,
- \( p \) is the probability of success on a single trial,
- \( 1 - p \) is the probability of failure on a single trial.
Substitute \( p = 0.0025 \) and \( n = 10 \) into the formula to compute the desired probability.
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