Problem 75 - Calculus for Biology and Medicine

Question

Suppose the waiting time for the first success in an experiment is geometrically distributed with mean \(1 / p\). (a) Find the probability that the first success occurs on the \(k\) th trial. (b) The experiment is repeated after the first success. Assume that the waiting time for the second success has the same distribution as the waiting time for the first success. Find the probability mass function for the distribution of the second success.

Step-by-Step Solution

Answer

(a) \( P(X = k) = (1-p)^{k-1} p \); (b) The distribution for the second success is the same: \( P(X = k) = (1-p)^{k-1} p \).

Step 1: Understand the Geometric Distribution

The geometric distribution models the number of trials until the first success in a series of independent Bernoulli trials, where each trial has the probability of success \( p \). The random variable \( X \) representing the waiting time for the first success follows the probability mass function \( P(X = k) = (1-p)^{k-1} p \). The mean of a geometric distribution is \( \frac{1}{p} \).

Step 2: Find the Probability of First Success on the k-th Trial

To find the probability that the first success occurs on the \( k \)-th trial, use the formula for the probability mass function of the geometric distribution. \[ P(X = k) = (1-p)^{k-1} p \] This formula directly provides the probability for the question asked.

Step 3: Define the Situation for the Second Success

The problem states that the experiment is repeated after the first success, with the waiting time for the second success having the same geometric distribution as the first success. Thus, the waiting time for each success follows the same geometric distribution with probability \( p \).

Step 4: Probability Mass Function for the Second Success

The probability that the second success occurs on the \( k \)-th trial after the first success can be modeled similarly, as each block of trials is independent and has the same distribution. Therefore, the probability mass function for the waiting time until the second success happening on the \( k \)-th trial after the first success is again:\[ P(X = k) = (1-p)^{k-1} p \] This demonstrates that each success, independent of the others, has the same geometric distribution.

Key Concepts

Probability Mass Function

In probability theory, the Probability Mass Function (PMF) describes the probabilities of a discrete random variable taking on specific values.
For the geometric distribution, which is often used to model the number of Bernoulli trials needed to obtain the first success, the PMF helps us calculate these probabilities.
The PMF for the geometric distribution is given by the formula:

• \[ P(X = k) = (1-p)^{k-1} p \]

Here,

• \( p \) is the probability of success in a single trial,
• \( k \) represents the trial number where the first success occurs.

This formula reflects the likelihood that the first success happens on the \( k \)-th trial, considering there were \( k-1 \) failures before it.
Understanding the PMF is essential, as it gives a comprehensive view of all possible outcomes of a random variable and their associated probabilities.

Bernoulli Trials

Bernoulli trials are the cornerstone of probability theory and play an essential role in defining the geometric distribution.
A Bernoulli trial is a simple experiment with exactly two outcomes: success or failure.
It is characterized by:

• Each trial has only two outcomes: "success" (with probability \( p \)) and "failure" (with probability \( 1-p \)).
• Every trial is identical and independent of others.

In the context of a geometric distribution, the series of Bernoulli trials continues until the first success occurs.
The geometric distribution then uses these trials to model waiting times for a success.
This is why understanding Bernoulli trials is crucial - they set the foundation for how probabilities are structured in more complex scenarios like geometric distributions.

Independent Trials

Understanding the concept of independent trials is vital in probability and statistics, particularly when dealing with geometric distribution.
Two trials are said to be independent if the outcome of one trial does not affect the outcome of another.
This implies:

• The probability of success \( p \) remains constant across all trials.
• The outcome of any trial does not alter the outcomes of future trials.

In the geometric distribution, independence is crucial because it ensures that the expected distribution of successes remains consistent, whether looking at the first, second, or further successes.
Each success or failure is unaffected by the sequence in which they occur, allowing the probability mass function to apply uniformly to each trial.
Grasping this concept helps in understanding why each block of Bernoulli trials in such distributions function similarly, maintaining the geometric distribution's integrity across independent events.