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Law: Bar Exam 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 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.

Step by step solution

01

Understanding the Distribution

The given problem involves a scenario where the number of attempts (\(n\)) continues until the first success (passing the exam). This is modeled using the geometric distribution. For a geometric distribution, the probability of success on each trial is \(p = 0.57\), and \(q = 1 - p = 0.43\).

02

Probability Distribution Formula

In a geometric distribution, the probability that the first success occurs on the \(n\)-th trial is given by the formula: \[ P(n) = q^{n-1} \times p \] where \(q = 0.43\) and \(p = 0.57\).

03

Probability for the Second Attempt

To find the probability that Bob first passes on his second attempt \((n=2)\), use the formula: \[ P(2) = q^{2-1} \times p = 0.43^1 \times 0.57 = 0.2451 \]

04

Probability for the Third Attempt

To find the probability that Bob first passes on his third attempt \((n=3)\), use the formula: \[ P(3) = q^{3-1} \times p = 0.43^2 \times 0.57 = 0.105363 \]

05

Probability for More Than Three Attempts

The probability that Bob needs more than three attempts can be calculated as the complement of passing within three attempts. First, calculate \(P(n > 3)\): \[ P(n > 3) = 1 - (P(1) + P(2) + P(3)) \] where \[ P(1) = 0.57 \], \[ P(2) = 0.2451 \], \[ P(3) = 0.105363 \]. \[ P(n > 3) = 1 - (0.57 + 0.2451 + 0.105363) = 0.079537 \]

06

Expected Number of Attempts

The expected number of attempts needed for the first success in a geometric distribution is \(\mu = \frac{1}{p}\). Here, \(p = 0.57\), so the expected number is \[ \mu = \frac{1}{0.57} \approx 1.754 \] rounding to three decimal places.

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Key Concepts

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

Probability

Probability is the chance or likelihood that a specific event will occur. In the case of the bar exam, probability helps us understand how likely Bob is to pass on a particular attempt. The core idea is to measure how often an event is expected to happen when considering the number of possible outcomes.
The formula for calculating the probability of passing on any given attempt using the geometric distribution is:

• First, identify the probability of success in one trial. For Bob, this probability is given as 57%, or 0.57.
• Next, determine the probability of all previous trials failing (until the success attempt). This is calculated as the success probability on one trial, raised to the number of failed attempts: \[ P(n) = q^{n-1} \times p \] where \( q = 0.43 \) and \( p = 0.57 \).

This formula provides a way to find out how likely it is that Bob will pass on any given try, such as the second or third attempt.

Random Variable

A random variable is a concept in probability that assigns a numerical value to each outcome of a random phenomenon. In our example with Bob and the bar exam, the random variable \( n \) represents the number of attempts Bob makes until he passes the exam.
Since \( n \) can take different values depending on when Bob passes, it's referred to as a discrete random variable with possible values \( n = 1, 2, 3, \ldots \).

• Random variables help in modeling and analyzing random events by providing a structured way to measure and express uncertainty.
• They facilitate the use of mathematical tools to calculate probabilities related to these random events, like the likelihood Bob passes on his first, second, or any successive attempt.

Ultimately, random variables are essential in transforming real-world scenarios into mathematical models that can be analyzed and utilized to make critical decisions.

Expected Value

Expected value, also known as the mean, provides a measure of the center of the probability distribution. For Bob's attempts at the bar which follows a geometric distribution, the expected value tells us how many attempts are needed on average before the first passing attempt.
The expected value of a geometric distribution is given by the formula:

• \( \mu = \frac{1}{p} \)

Calculating for Bob, where \( p = 0.57 \), we find:

• \( \mu = \frac{1}{0.57} \approx 1.754 \)

This suggests that, on average, Bob is expected to pass the exam in about 1.754 attempts.
This insight is crucial for Bob to plan his study schedule and manage his expectations about the process of passing the bar exam.

Bar Exam

The bar exam is a significant milestone for law graduates like Bob, determining their ability to practice law. Passing this exam is essential, and understanding the statistics behind it can greatly influence Bob's strategies in preparation.

• The bar exam typically tests a candidate's knowledge and understanding of the legal system, encompassing numerous legal principles and applications.
• With a pass rate of about 57%, it's important for test-takers to consider both their preparation and psychological readiness.
• Geometric probability allows candidates like Bob to understand their odds of passing and thereby support strategic studying and retake planning if necessary.

Let's remember that each attempt is not just an isolated event but a step closer to understanding the broader legal field.