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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.

Short Answer

Expert verified
(a) \( P(n) = 0.23^{n-1} \times 0.77 \). (b) 0.77. (c) 0.1771. (d) 0.0529. (e) 1.30 attempts.

Step by step solution

01

Understand the Geometric Distribution

The problem involves a situation where we perform a series of independent trials until a success occurs. This is a classic case of a geometric distribution, where the probability of success in one trial, \( p \), is given as \( 77\% \) or \( 0.77 \).
02

Probability Distribution Formula

For a geometric distribution, the probability that the first success occurs on trial \( n \) is given by \( P(n) = (1-p)^{n-1} \times p \). For this problem, it is \( P(n) = 0.23^{n-1} \times 0.77 \).
03

Probability of Passing on First Try

We need to calculate \( P(n = 1) \). Using the formula from Step 2, substitute \( n = 1 \): \( P(1) = 0.23^{1-1} \times 0.77 = 0.77 \).
04

Probability of Passing on Second Try

Calculate \( P(n = 2) \). Using the formula \( P(2) = 0.23^{2-1} \times 0.77 = 0.23 \times 0.77 = 0.1771 \).
05

Probability of Needing Three or More Tries

The probability of needing three or more tries is the complement of the probability of passing in one or two tries: \( 1 - P(n = 1) - P(n = 2) = 1 - 0.77 - 0.1771 = 0.0529 \).
06

Expected Number of Attempts

For a geometric distribution, the expected value \( \mu \) is \( \mu = \dfrac{1}{p} \). Thus, \( \mu = \dfrac{1}{0.77} \approx 1.30 \).

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Key Concepts

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

Probability Distribution
A geometric distribution is a specific type of probability distribution used to model the number of trials it takes to achieve a first success. In probability terms, a trial refers to an event with two possible outcomes: "success" and "failure." The probability of success during each trial is symbolized by \( p \).
In our example of Susan taking Western Civilization, we assign the probability \( p = 0.77 \) (reflecting the 77% pass rate). This distribution tells us the likelihood of Susan passing on her first, second, or any subsequent attempt.
For a geometric distribution, we calculate the probability of the first success happening on trial \( n \) (e.g., passing on the first, second try, etc.) using the formula:
  • \( P(n) = (1-p)^{n-1} \times p \)
Here, \( (1-p)^{n-1} \) represents the probability of failing the first \( n-1 \) trials and \( p \) is the probability that the \( n^{th} \) trial is a success. This formula is central to determining chances in each trial sequence, assisting in predicting when a specific outcome like Susan's first pass will happen.
Expected Value
The expected value is a way of determining the average outcome of a probability distribution over the long run. In a geometric distribution, the expected value \( \mu \) is derived using the formula:
  • \( \mu = \dfrac{1}{p} \)
This formula tells us the mean number of trials needed for the first success.
In Susan's context, the expected value represents the typical number of attempts required for her to pass her Western Civilization course on the first try. Since her pass probability is 0.77, we calculate her expected number of attempts as:
  • \( \mu = \dfrac{1}{0.77} \approx 1.30 \)
The expected value of 1.30 signifies that, on average, it would take Susan just over one attempt to pass the course. Understanding \( \mu \) provides valuable insight into the efficiency or difficulty of completing a task analyzing potential outcomes with average scenarios.
Independent Trials
In the context of probability distributions, independent trials are crucial for analyzing geometric distributions. Each trial has no impact on the other trials, meaning the probability of success or failure remains constant throughout each trial.
For Susan, each attempt at her Western Civilization course is considered an independent trial. Whether she fails or passes on one attempt does not change her chances for subsequent attempts. It's like flipping a fair coin where each flip is independent of the others.
The concept of independence underlines why the probability of success, \( p = 0.77 \), is applied uniformly across all attempts. It allows us to use the geometric distribution formula effectively. Recognizing trials as independent ensures the model accurately represents the statistical process and allows reliable predictions based on historical data.

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Most popular questions from this chapter

Bob is a recent law school graduate who intends to take the state bar exam. According to the National Conference on Bar Examiners, about \(57 \%\) of all people who take the state bar exam pass (Source: The Book of \(\mathrm{Odds}\) by Shook and Shook, Signet). Let \(n=1,2,3, \ldots\) represent the number of times a person takes the bar exam until the first pass. (a) Write out a formula for the probability distribution of the random variable \(n\). (b) What is the probability that Bob first passes the bar exam on the second try \((n=2) ?\) (c) What is the probability that Bob needs three attempts to pass the bar exam? (d) What is the probability that Bob needs more than three attempts to pass the bar exam? (e) What is the expected number of attempts at the state bar exam Bob must make for his (first) pass? Hint: Use \(\mu\) for the geometric distribution and round.

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