91Ó°ÊÓ

College: Core Requirement Susan is taking western civilization this semester on a pass/fail basis. The department teaching the course has a history of passing \(77 \%\) of the students in western civilization each term. Let \(n=1,2,3, \ldots\) represent the number of times a student takes western civilization until the first passing grade is received. (Assume the trials are independent.) (a) Write out a formula for the probability distribution of the random variable \(n\). (b) What is the probability that Susan passes on the first try \((n=1) ?\) (c) What is the probability that Susan first passes on the second try \((n=2) ?\) (d) What is the probability that Susan needs three or more tries to pass western civilization? (e) What is the expected number of attempts at western civilization Susan must make to have her (first) pass? Hint: Use \(\mu\) for the geometric distribution and round.

Step by step solution

01

Identify the Distribution

The problem describes a scenario fitting a geometric distribution: the number of trials until the first success (pass). In this case, a 'success' is defined as passing the course.

02

Write the Probability Formula

For a geometric distribution with success probability \(p\), the probability that the first success occurs on the \(n^{th}\) trial is given by:\[ P(n) = (1-p)^{n-1} \, p \]Here, \(p = 0.77\), so the probability formula becomes:\[ P(n) = (1-0.77)^{n-1} \times 0.77 = 0.23^{n-1} \times 0.77 \]

03

Calculate Probability for First Attempt (n=1)

For \(n = 1\), the probability that Susan passes on the first attempt is:\[ P(1) = 0.23^{1-1} \times 0.77 = 1 \times 0.77 = 0.77 \]

04

Calculate Probability for Second Attempt (n=2)

For \(n = 2\), the probability that the first success is on the second try is:\[ P(2) = 0.23^{2-1} \times 0.77 = 0.23 \times 0.77 = 0.1771 \]

05

Calculate Probability for Three or More Tries

The probability that it takes three or more tries to pass is the complement of passing on the first or second try. It is given by:\[ P(n \geq 3) = 1 - P(n=1) - P(n=2) \]Substituting the values calculated earlier:\[ P(\geq 3) = 1 - 0.77 - 0.1771 = 0.0529 \]

06

Find the Expected Number of Attempts

The expected number of attempts for a geometric distribution is given by the formula \(\mu = \frac{1}{p}\), where \(p = 0.77\):\[ \mu = \frac{1}{0.77} \approx 1.2987 \]So Susan should expect to attempt the course approximately 1.3 times.

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Key Concepts

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

Probability Distribution

In this exercise, we see an example of a geometric probability distribution. A probability distribution, at its core, explains the likelihood of each outcome in a sample space. For the geometric distribution, the outcome of interest here is about how long it will take until the first success occurs - in this case, until Susan passes her course.

The number of trials, denoted as \(n\), represents how many attempts are needed for the first success. The formula for the probability of achieving the first success on the \(n^{th}\) trial is:

• \[ P(n) = (1-p)^{n-1} \, p \]

Here, \( (1-p)^{n-1} \) represents the probability of failing the previous \(n-1\) tries, while \(p\) is the probability of success on the \(n^{th}\) try. It is crucial to understand this distribution because it provides insights into how random phenomena behave over several trials.

Expected Value

The expected value is a powerful concept that tells us the average number of trials expected to achieve the first success in a probability distribution. For a geometric distribution, the expected value, denoted as \( \mu \), is given by:

• \[ \mu = \frac{1}{p} \]

This formula arises because each trial in the geometric distribution is independent, and each success has a constant probability \(p\).

In Susan's case, with the probability of passing being \(0.77\), the expected number of attempts to pass is approximately \(1.3\). This doesn’t mean Susan will always pass in 1.3 trials; rather, it means that on average, over many students taking the course, the first pass happens after around 1.3 attempts. The expected value gives us a statistical landmark, pointing out what happens typically over numerous realizations of the same scenario.

Independent Trials

Independent trials are a significant aspect of a geometric distribution, ensuring that the outcome of one trial does not impact another. In the context of Susan attempting the course, each try is independent because whether she passes in one attempt, doesn't alter the probability of passing the next round.

To break it down further:

• The probability of passing remains constant at \(0.77\) for every attempt.
• The trials do not "remember" previous outcomes.
• This independency ensures the use of formulas like \(P(n) = (1-p)^{n-1} p\), and \(\mu = \frac{1}{p}\).

Understanding that each attempt is independent is crucial, as it reassures that the calculated probabilities and expected values remain true regardless of past results. Consistent quality and reliance on independence make prediction models like these valuable for realistic academic and professional settings.