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Suppose that the random variable \(X\) has a geometric distribution with \(p=0.5 .\) Determine the following probabilities: a. \(P(X=1)\) b. \(P(X=4)\) c. \(P(X=8)\) d. \(P(X \leq 2)\) e. \(P(X>2)\)

Short Answer

Expert verified
a. 0.5, b. 0.0625, c. 0.00390625, d. 0.75, e. 0.25

Step by step solution

01

Understanding the Geometric Distribution

The geometric distribution models the number of trials needed to get the first success in a series of independent and identically distributed Bernoulli trials. The probability mass function (PMF) is given by \(P(X = k) = (1-p)^{k-1} \cdot p\), where \(p\) is the probability of success on each trial and \(k\) is the trial number on which the first success occurs.
02

Calculating P(X=1)

Use the probability mass function with \(k = 1\) and \(p=0.5\). Substitute these values into the PMF: \(P(X = 1) = (1-0.5)^{1-1} \times 0.5 = 1 \times 0.5 = 0.5\).
03

Calculating P(X=4)

Use the PMF formula with \(k = 4\): \(P(X = 4) = (1-0.5)^{4-1} \times 0.5 = 0.5^3 \times 0.5 = 0.5^4 = 0.0625\).
04

Calculating P(X=8)

Apply the PMF formula with \(k = 8\): \(P(X = 8) = (1-0.5)^{8-1} \times 0.5 = 0.5^7 \times 0.5 = 0.5^8 = 0.00390625\).
05

Calculating P(X≤2)

This probability is the sum of \(P(X=1)\) and \(P(X=2)\). Calculate \(P(X=2)\) using the PMF: \(P(X=2) = (1-0.5)^{2-1} \times 0.5 = 0.5^1 \times 0.5 = 0.25\). Thus, \(P(X\leq 2) = P(X=1) + P(X=2) = 0.5 + 0.25 = 0.75\).
06

Calculating P(X>2)

Recognize that \(P(X>2) = 1 - P(X\leq 2)\). We've already calculated \(P(X\leq 2) = 0.75\). Thus, \(P(X>2) = 1 - 0.75 = 0.25\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Mass Function
To comprehend the geometric distribution, it is essential to understand the concept of a Probability Mass Function (PMF). In the case of a geometric distribution, the PMF is used to find the probability that the first success occurs on the k-th trial. When we talk about a PMF, we essentially mean a function that provides the probability of each possible discrete outcome of a random variable. In the context of the geometric distribution, our random variable is the number of trials needed for the first success. For a geometric distribution, the PMF is represented mathematically as:
  • \( P(X = k) = (1-p)^{k-1} \cdot p \)
Here, \( p \) is the probability of success for each independent trial. The term \((1-p)^{k-1}\) signifies the probability of having \(k-1\) failures before the first success occurs on the k-th trial. This equation is the foundation of understanding probabilities in geometric distributions, as it tells us how likely it is to encounter our first success after several attempts.
Calculating Probabilities
When working with a geometric distribution, calculating probabilities involves using the PMF to determine the likelihood of certain events. Let's go through a few examples to see how this works in practice:
  • Example 1: To calculate \( P(X = 1) \), you substitute \( k = 1 \) and \( p = 0.5 \) into the PMF formula. This gives \( P(X = 1) = (1-0.5)^{1-1} \times 0.5 = 0.5 \).
  • Example 2: For \( P(X = 4) \), use \( k = 4 \) and \( p = 0.5 \). The calculation looks like this: \( P(X = 4) = (1-0.5)^{4-1} \times 0.5 = 0.0625 \).
  • Summary of steps: Plug the values into the formula, solve the equation for each scenario, and interpret the results to understand the probability of specific outcomes.
Having a clear grasp of these calculations is crucial for mastering geometric distributions, as it allows us to solve various problems involving probabilities of first successes in trials.
Bernoulli Trials
Understanding Bernoulli trials is key to grasping the concept of a geometric distribution since the entire premise is built upon these trials. A Bernoulli trial can be defined simply:
  • It is an experiment or process that results in a single outcome.
  • There are only two possible results: success or failure.
  • Each trial is independent of others.
  • The probability of success, \( p \), remains constant in every trial.
In the context of geometric distribution, the outcome of interest is the trial number when the first success occurs. Thus, a series of Bernoulli trials can be viewed as the groundwork for events modeled by a geometric random variable.
Geometric distributions specifically deal with the wait time, or the number of trials, to see the first success under these conditions. Hence, through this understanding, it becomes easier to analyze and predict outcomes using the tools and formulas provided by geometric distributions.

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