91Ó°ÊÓ

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 \(\operatorname{try}(n=2) ?\) (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 Geometric Distribution

This problem involves a geometric distribution since it describes the probability of 'failure' until the first 'success'. For the geometric distribution, the probability that the first success occurs on the nth trial is given by the formula:\[ P(n) = (1-p)^{n-1} \cdot p \]where \(p\) is the probability of success on each trial.

02

Calculating Probability for n=2

We need to calculate the probability that Bob first passes the bar exam on the second try \( (n=2) \). Using the formula, we substitute \(n=2\) and \(p=0.57\):\[ P(2) = (1-0.57)^{2-1} \cdot 0.57 = 0.43 \cdot 0.57 = 0.2451 \]

03

Calculating Probability for n=3

Next, find the probability that Bob first passes on his third attempt \( (n=3) \). Using the formula:\[ P(3) = (1-0.57)^{3-1} \cdot 0.57 = 0.43^{2} \cdot 0.57 = 0.1845 \]

04

Calculating Probability for More Than Three Attempts

The probability that Bob needs more than three attempts means the success does not occur on the first, second, or third attempts. This can be calculated as:\[ P(n > 3) = 1 - (P(1) + P(2) + P(3)) \]Calculate \(P(1)\):\[ P(1) = (1-0.57)^{1-1} \cdot 0.57 = 0.57 \]Thus,\[ P(n > 3) = 1 - (0.57 + 0.2451 + 0.1845) = 1 - 1.0006 \approx 0 \]

05

Calculating the Expected Number of Attempts

The expected number of attempts for a geometric distribution is given by:\[ E(n) = \frac{1}{p} \]Substituting \(p = 0.57\), we find:\[ E(n) = \frac{1}{0.57} \approx 1.754 \]Since we are asked to round, it's better understood that Bob is expected to pass in about 2 attempts.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Probability Distribution

The concept of a probability distribution is essential when dealing with random experiments, like taking a bar exam multiple times. It provides a mathematical function that describes the likelihood of possible outcomes. For the geometric distribution, the probability distribution shows the probability that the first success (passing the exam) occurs at a specific trial (attempt).

In a geometric distribution, we use the formula:

• \[ P(n) = (1-p)^{n-1} \cdot p \]

Where:

• \( n \) is the trial number on which the first success occurs.
• \( p \) is the probability of success on each attempt (57% for passing the bar exam).
• \( 1-p \) symbolizes the probability of failure on each trial. For Bob, it's 43%.

The outcomes like passing after the 1st, 2nd, or 3rd try can be calculated using this formula, producing a distribution of probabilities for each potential attempt. Understanding this distribution helps in predicting the various scenarios Bob might face.

Expected Value

The expected value in probability gives us the average outcome we expect from a random variable over many trials. It's like asking how many times, on average, Bob will need to take the bar exam to pass, considering the probability of success.

• For the geometric distribution, the expected value formula is:\[ E(n) = \frac{1}{p} \]Here, \( p \) is the probability of success. In this case, 0.57.

Using Bob's situation:

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

This indicates that Bob is expected to pass the bar exam in approximately 1.754 attempts or, realistically, about 2 attempts when rounded to the nearest whole number. The expected value does not guarantee that he will pass on the second try, but it gives a statistical yardstick for planning and expectations.

Random Variable

A random variable is a numerical description of the outcome of a random phenomenon. In the context of Bob taking the bar exam, the random variable \( n \) represents the number of attempts it takes until he passes the exam.

Understanding random variables is crucial as they help translate real-life, random scenarios into a format we can analyze mathematically. In Bob's scenario:

• The potential values of \( n \) are \( 1, 2, 3, \ldots \), representing each attempt.
• Each value of \( n \) is associated with a probability that can be calculated using the geometric distribution formula.

By analyzing this random variable, we can predict outcomes like passing on the second attempt, third attempt, or taking more than three attempts. Converting real-world problems into random variables allows us to use mathematical tools to derive meaningful insights from seemingly unpredictable events.