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 \(\mathrm{Odds}\) 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.

Step by step solution

01

Define the Geometric Distribution

The probability distribution for the number of trials needed for the first success in a series of Bernoulli trials (like passing or failing an exam) follows a geometric distribution. Here, the probability of success for each trial is \( p = 0.57 \). The probability mass function (PMF) for a geometric distribution is \( P(n) = (1-p)^{n-1} \cdot p \). In this scenario, \( (1-p) = 0.43 \) is the probability of failing the trial.

02

Formula for the Probability Distribution

Using the geometric PMF, we write the formula for the distribution as \( P(n) = 0.43^{n-1} \times 0.57 \). This formula gives the probability that Bob passes the exam on the \( n \)-th attempt.

03

Probability of Passing on the Second Try

To find the probability that Bob passes on his second attempt \((n=2)\), use the PMF in Step 2: \( P(2) = 0.43^{2-1} \times 0.57 = 0.43 \times 0.57 = 0.2451 \).

04

Probability of Passing on the Third Try

For the third attempt \((n=3)\), compute using the formula: \( P(3) = 0.43^{3-1} \times 0.57 = 0.43^2 \times 0.57 = 0.185022 \).

05

Probability of More Than Three Attempts

The probability that Bob passes on more than the third attempt is \(1 - (P(1) + P(2) + P(3))\). Calculate these terms as: \( P(1) = 0.57 \), \( P(2) = 0.2451 \), and \( P(3) = 0.185022 \).Therefore, \( P(n > 3) = 1 - (0.57 + 0.2451 + 0.185022) = 1 - 0.998122 = 0.001878 \).

06

Calculate Expected Number of Attempts

The expected number of attempts for a geometric distribution is given by the formula \( \mu = \frac{1}{p} \). Substituting \( p = 0.57 \), we get \( \mu = \frac{1}{0.57} \approx 1.7544 \).

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Key Concepts

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

Probability Distribution

A probability distribution helps us understand how probabilities are distributed across different outcomes. When discussing the number of attempts Bob makes before passing the exam, each possible attempt can be seen as an outcome with its dedicated probability. In this case, the scenario follows a geometric distribution, which is a type of probability distribution frequently encountered in situations with repeated trials until the first success.

Whenever dealing with geometric distributions, the focus is on the probability of the first success occurrence after the event having failed multiple times. The basic structure here involves figuring out the probability of failure for certain attempts before achieving success. This knowledge is vital in predicting future occurrences and understanding the likelihood of various outcomes.

Bernoulli Trials

Bernoulli trials are the fascinating foundation for understanding how each of Bob’s attempts is independently considered a trial. In these trials, there are only two possible outcomes: success or failure. For each attempt, either Bob passes or fails the exam, thus simplifying the scenario into a Bernoulli trial situation.

What's intriguing about Bernoulli trials is that each trial is independent of the others. That means Bob’s attempts to pass the bar exam don't influence each other. The probability of passing stays constant at 57% for each attempt, making it a perfect candidate for geometric distribution analysis. By grasping this, students can appreciate the underlying mechanics of probability calculations based on repeated independent events.

Expected Value

The expected value is an essential concept for predicting an average outcome over numerous trials or attempts. For the problem at hand, the expected value represents the average number of attempts Bob might need to pass the bar exam by calculating it through the geometric distribution formula.

To find this expected value, use the formula \( \mu = \frac{1}{p} \), where \( p \) is the probability of success on each trial (57% in Bob’s case). Thus, \( \mu \approx 1.7544 \). This computed mean, around 1.7544, hypothetically tells us the average amount of attempts you'd expect Bob might undergo to pass. Understanding expected value empowers students to see beyond just immediate likelihoods and reason about longer-term averages.

Probability Mass Function

A probability mass function (PMF) provides the precise probability of each individual outcome. For a geometric distribution like Bob's situation with the bar exam, the PMF helps calculate the likelihood of passing the test on a specific attempt.

The PMF for a geometric distribution can be expressed as \( P(n) = (1-p)^{n-1} \cdot p \). In Bob’s case, \( (1-p) = 0.43 \), reflecting the chance he fails a single trial. Using the PMF, various probabilities are computed for each attempt number. For instance, the chance of passing on the second attempt is \( P(2) = 0.2451 \), while for the third attempt, it’s \( P(3) = 0.185022 \). By leveraging the PMF, detailed insights into potential outcomes and their likelihoods become accessible, clarifying the broader concept of probability distribution in a geometric setting.