Problem 45 - A First Course in Probability

Question

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or an "off" day. He figures that if he has an on day, then each of his examiners will pass him, independently of each other, with probability \(.8,\) whereas if he has an off day, this probability will be reduced to \(.4 .\) Suppose that the student will pass the examination if a majority of the examiners pass him. If the student feels that he is twice as likely to have an off day as he is to have an on day, should he request an examination with 3 examiners or with 5 examiners?

Step-by-Step Solution

Answer

The student should request an examination with 5 examiners, as the overall passing probability with 5 examiners is slightly higher (\(0.538\)) than with 3 examiners (\(0.533\)).

Step 1: Define probabilities

First, let's define the probabilities given in the problem. Let's denote: 1. \(P(O)\) - The probability of having an "on" day, 2. \(P(OF)\) - The probability of having an "off" day, 3. \(P(P_E|O)\) - The probability of passing an individual examiner on an "on" day, 4. \(P(P_E|OF)\) - The probability of passing an individual examiner on an "off" day. We are given: 1. \(P(P_E|O) = 0.8\), 2. \(P(P_E|OF) = 0.4\), 3. The student is twice as likely to have an "off" day than an "on" day, which means: \[P(OF) = 2 * P(O)\] Since \(P(O) + P(OF) = 1\), we can find the values of \(P(O)\) and \(P(OF)\).

Step 2: Calculate P(O) and P(OF)

Using the information given, we can calculate the probabilities of having an "on" day and an "off" day as follows: \[P(O) + P(OF) = 1\] \[P(O) + 2 * P(O) = 1\] \[3 * P(O) = 1\] \[P(O) = \frac{1}{3}\] Thus, we have: \[P(OF) = 1 - P(O) = 1 - \frac{1}{3} = \frac{2}{3}\]

Step 3: Calculate passing probabilities

Now, we calculate the probability of passing the examination with 3 examiners and with 5 examiners on both "on" and "off" days. The student will pass the exam if a majority of the examiners pass him. For 3 examiners, he must pass at least 2 examiners, and for 5 examiners, he must pass at least 3 examiners. We can calculate these probabilities using binomial probabilities: \[P(P_3|O) = {3 \choose 2} * 0.8^2 * 0.2^1 + {3 \choose 3} * 0.8^3 * 0.2^0 = 0.384 + 0.512 = 0.896\] \[P(P_3|OF) = {3 \choose 2} * 0.4^2 * 0.6^1 + {3 \choose 3} * 0.4^3 * 0.6^0 = 0.288 + 0.064 = 0.352\] \[P(P_5|O) = {5 \choose 3} * 0.8^3 * 0.2^2 + {5 \choose 4} * 0.8^4 * 0.2^1 + {5 \choose 5} * 0.8^5 * 0.2^0 = 0.4096 + 0.32768 + 0.032768 = 0.98\] \[P(P_5|OF) = {5 \choose 3} * 0.4^3 * 0.6^2 + {5 \choose 4} * 0.4^4 * 0.6^1 + {5 \choose 5} * 0.4^5 * 0.6^0 = 0.2304 + 0.0768 + 0.01024 = 0.31744\]

Step 4: Calculate overall passing probabilities

Now, we can calculate the overall passing probabilities for both 3 examiners and 5 examiners using the "on" and "off" day probabilities: \[P(P_3) = P(P_3|O) * P(O) + P(P_3|OF) * P(OF) = 0.896 * \frac{1}{3} + 0.352 * \frac{2}{3} \approx 0.533\] \[P(P_5) = P(P_5|O) * P(O) + P(P_5|OF) * P(OF) = 0.98 * \frac{1}{3} + 0.31744 * \frac{2}{3} \approx 0.538\]

Step 5: Compare the probabilities and decide

Comparing the overall passing probabilities for 3 examiners and 5 examiners: \[P(P_3) \approx 0.533\] \[P(P_5) \approx 0.538\] Since the probability of passing with 5 examiners is slightly higher than with 3 examiners (\(0.538 > 0.533\)), the student should request an examination with 5 examiners.

Key Concepts

Understanding Binomial Probability

When preparing for an important examination, students often grapple with the question of what factors will influence their chances of success. An insightful way to model such scenarios is through binomial probability.

Binomial probability refers to the probability of a specific number of successes in a series of independent trials, each with the same probability of success. In the context of examinations, imagine assessing the likelihood of a student passing individual examiners on a 'good' or 'bad' day. This approach is particularly powerful because it provides a clear formula for calculating scenarios where each success (or failure) is independent of another.

Using the binomial probability formula:

• Binomial Probability Formula:

• Let's consider an important aspect of the student's strategy. For a test with 3 examiners, the student would need to pass at least 2 examiners. Using binomial distribution, the probabilities can be summed as the chances of passing exactly 2 examiners, plus the probability of passing all 3. The same logic applies for 5 examiners, passing at least 3. This leads to the calculations from the original exercise for various scenarios involving 'on' or 'off' days.
  
  By understanding the fundamentals of binomial probability, students can make informed decisions that statistically improve their chances of passing an exam.

Grasping Conditional Probability in Exams

Within the realm of probabilities in examinations, conditional probability plays a vital role. It allows students to calculate the probability of an event occurring given that another event has already occurred.

Applying this to the student's situation, the concept of conditional probability helps us understand the likelihood of passing each examiner conditioned on whether the student is having an 'on' day or an 'off' day. Specifically, we examine the probability of passing an examiner (event A) given that the student is on an 'on' day (condition B), expressed as P(A|B).

The key takeaway here is recognizing that the probability of passing isn't uniform; it changes based on the student's condition (i.e., the type of day they are experiencing). It's these dynamic probabilities that make conditional probability an essential concept, influencing not only the outcome of the student's exam day but also the decision on the number of examiners to request for the examination.

Decoding the Majority Voting System

Exam strategies are not just about preparation; they also encompass decision-making, like deciding how many examiners one should face in a viva voce. This is where the 'majority voting system' concept comes into play—a decision-making process often used in group settings, including exams.

A majority vote system implies that to succeed, one needs more than half the votes—in this case, passes from the examiners. For the student in our exercise, this means having more examiners passing than failing him. It skews the probability and requires a significant consideration of both binomial probability and conditional probability in decision-making.

In the context of having 3 versus 5 examiners, intuitively, one might think more examiners could mean more opportunities to fail. However, by diving into the probabilities, yielding a deeper understanding of these votes, it's revealed that a greater number of examiners actually increases the chances of success due to the leniency in the required 'majority'. As the final step of the solution implies, with 5 examiners, the slight rise in passing probability results from the interplay between binomial distribution and the majority voting dynamics. In essence, the majority voting system adds a layer to our probability calculations, one that may sway the odds in the student's favor.