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Chapter 6: Problem 19

A multiple choice exam is given. A problem has four possible answers, and exactly one answer is correct. The student is allowed to choose a subset of the four possible answers as his answer. If his chosen subset contains the correct answer, the student receives three points, but he loses one point for each wrong answer in his chosen subset. Show that if he just guesses a subset uniformly and randomly his expected score is zero.

Short Answer

Expert verified
The expected score is zero when guessing a subset uniformly and randomly.

Step by step solution

01

Define Possible Subsets

The student can choose a subset from the four possible answers. The total possible subsets are all the combinations of these four answers, which is \(2^4 = 16\). These include the empty set, single answer sets, two-answer sets, three-answer sets, and the full set of four answers.
02

Calculate Points for Each Subset

For any chosen subset, if it contains the correct answer, the student starts with 3 points. However, for every incorrect answer in the subset, the student loses one point. Therefore, the points for a subset with \(k\) elements containing the correct answer is \(3 - (k-1)\), as \(k-1\) represents the number of incorrect choices.
03

Calculate Expected Score for Each Subset Size

We need to compute the expected score for subsets of each size. For a subset of size \(k\):- Probability that the subset includes the correct answer = \(\frac{k}{4}\).- Expected score = \( \frac{k}{4} imes (3 - (k-1)) + (1 - \frac{k}{4}) imes (-k) \).
04

Simplify the Expected Score Expression

Simplify the expected score expression obtained in Step 3. This simplifies to:\[ E = \frac{k}{4} \times (4 - k) - (1 - \frac{k}{4}) \times k \]\[ E = \frac{k(4 - k) - (4 - k)k}{4} = 0 \]This shows that the expected score is 0 for any \(k\) sized subset.
05

Conclude the Expected Score for Random Subset Choice

Since the expected score for each possible subset size is zero, when guessing a subset uniformly and randomly, the overall expected score remains zero, fulfilling the conditions of the problem.

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Key Concepts

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

Probability Theory
Probability theory is the mathematical framework that helps us analyze events, especially when dealing with uncertainty. It is essential in various fields like finance, science, and everyday decision-making. In the given problem, probability theory is used to determine the likelihood of a student choosing a correct subset of answers in a multiple-choice exam.
  • This involves calculating the probability of picking subsets that include the correct answer among many possible choices.
  • By understanding these probabilities, we can make informed predictions about expected outcomes, such as the student's expected score.
Here, probability theory helps us conclude that the student breaks even when guessing randomly, as all possible outcomes are accounted for in the expected calculation.
Random Variables
Random variables are used to quantify the outcomes of random processes. They can take on different values based on chance, and these values are determined by a probability distribution. In our exercise, if we represent each possible subset choice as a random variable, we can calculate its expected value, helping us understand what to expect on average when a student randomly chooses a subset of answers.
  • The random variable here represents the score the student gets from choosing a particular subset.
  • The possible values of this random variable range from negative scores (penalties for incorrect answers) to positive scores (reward for the correct answer).
The expected value of the random variable gives insights into the average outcome, which proves to be zero when guessing randomly.
Combinatorics
Combinatorics is the branch of mathematics that deals with counting and arranging objects. It is crucial in determining the number of possible subsets of answers a student can choose from in the exercise. The problem involves creating subsets of a set that contains four possible answers, with the combinatorial aspect calculating the total number of subsets, which is expressed as powers of two.
  • The number of possible subsets, including all sizes from empty to full sets, is generated by the formula for subsets: \(2^n\).
  • For four possible answers, the resulting number of subsets is \(2^4 = 16\).
Understanding combinatorics allows us to systematically evaluate all possible choices the student can make, later affecting the expected value calculation.
Probability Distribution
A probability distribution describes how probabilities are assigned to all possible outcomes of a random event. In this exercise, the distribution is equal because the student is choosing a subset randomly and uniformly from the possible answers. Each subset has an equal likelihood of being selected.
  • The uniform distribution implies an equal probability for each subset, ensuring every subset is equally likely to contain the correct answer, thus affecting the scoring system.
  • For subsets of size \(k\), the uniform distribution probability helps in calculating whether that subset likely contains the correct answer.
With a balanced distribution, it aids in proving why the expected value of the student's score averages out to zero, aligning with the randomness of their guessing strategy.

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Most popular questions from this chapter

Let \(X\) be a random variable distributed uniformly over \([0,20] .\) Define a new random variable \(Y\) by \(Y=\lfloor X\rfloor\) (the greatest integer in \(X) .\) Find the expected value of \(Y\). Do the same for \(Z=\lfloor X+.5\rfloor\). Compute \(E(|X-Y|)\) and \(E(|X-Z|) . \quad\) (Note that \(Y\) is the value of \(X\) rounded off to the nearest smallest integer, while \(Z\) is the value of \(X\) rounded off to the nearest integer. Which method of rounding off is better? Why?)

A deck of ESP cards consists of 20 cards each of two types: say ten stars, ten circles (normally there are five types). The deck is shuffled and the cards turned up one at a time. You, the alleged percipient, are to name the symbol on each card before it is turned up. Suppose that you are really just guessing at the cards. If you do not get to see each card after you have made your guess, then it is easy to calculate the expected number of correct guesses, namely ten. If, on the other hand, you are guessing with information, that is, if you see each card after your guess, then, of course, you might expect to get a higher score. This is indeed the case, but calculating the correct expectation is no longer easy. But it is easy to do a computer simulation of this guessing with information, so we can get a good idea of the expectation by simulation. (This is similar to the way that skilled blackjack players make blackjack into a favorable game by observing the cards that have already been played. See Exercise 29.) (a) First, do a simulation of guessing without information, repeating the experiment at least 1000 times. Estimate the expected number of correct answers and compare your result with the theoretical expectation. (b) What is the best strategy for guessing with information? (c) Do a simulation of guessing with information, using the strategy in (b). Repeat the experiment at least 1000 times, and estimate the expectation in this case. (d) Let \(S\) be the number of stars and \(C\) the number of circles in the deck. Let \(h(S, C)\) be the expected winnings using the optimal guessing strategy in (b). Show that \(h(S, C)\) satisfies the recursion relation $$h(S, C)=\frac{S}{S+C} h(S-1, C)+\frac{C}{S+C} h(S, C-1)+\frac{\max (S, C)}{S+C}$$ and \(h(0,0)=h(-1,0)=h(0,-1)=0 .\) Using this relation, write a program to compute \(h(S, C)\) and find \(h(10,10)\). Compare the computed value of \(h(10,10)\) with the result of your simulation in (c). For more about this exercise and Exercise 26 see Diaconis and Graham. \(^{11}\)

An insurance company has 1,000 policies on men of age \(50 .\) The company estimates that the probability that a man of age 50 dies within a year is .01 . Estimate the number of claims that the company can expect from beneficiaries of these men within a year.

A random sample of 2400 people are asked if they favor a government proposal to develop new nuclear power plants. If 40 percent of the people in the country are in favor of this proposal, find the expected value and the standard deviation for the number \(S_{2400}\) of people in the sample who favored the proposal.

A card is drawn at random from a deck consisting of cards numbered 2 through 10. A player wins 1 dollar if the number on the card is odd and loses 1 dollar if the number if even. What is the expected value of his winnings?
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