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Chapter 3: Problem 28

Edison Research gathered exit poll results from several sources for the Wisconsin recall election of Scott Walker. They found that \(53 \%\) of the respondents voted in favor of Scott Walker. Additionally, they estimated that of those who did vote in favor for Scott Walker, \(37 \%\) had a college degree, while \(44 \%\) of those who voted against Scott Walker had a college degree. Suppose we randomly sampled a person who participated in the exit poll and found that he had a college degree. What is the probability that he voted in favor of Scott Walker? \(^{47}\)

Short Answer

Expert verified
The probability that a college graduate voted for Walker is approximately 48.64%.

Step by step solution

01

Identify Given Information

We know that 53% of respondents voted in favor of Scott Walker. Of those, 37% had a college degree, and 44% of those voting against him had a college degree. We need to find the probability that a person with a college degree voted for Scott Walker.
02

Define Events

Let \(A\) be the event that a person voted for Scott Walker, and \(B\) the event that a person has a college degree. We are tasked with finding \( P(A|B) \), the probability that a person voted for Walker, given that they have a college degree.
03

Apply Bayes' Theorem

We will use Bayes' Theorem: \( P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} \). We have to calculate \(P(B|A)\), the probability of having a college degree if the person voted for Walker, \(P(A)\), the probability of voting for Walker, and \(P(B)\), the probability of having a college degree.
04

Calculate P(B|A) and P(A)

We know \(P(B|A) = 0.37\) (37% of Walker's voters have a college degree) and \(P(A) = 0.53\) (53% voted for Walker).
05

Calculate P(B)

To calculate \(P(B)\), we use the law of total probability: \(P(B) = P(B|A)P(A) + P(B|A^c)P(A^c)\). We know \(P(B|A^c) = 0.44\) (44% of non-Walker voters with a college degree) and \(P(A^c) = 0.47\) (47% voted against Walker). So, \(P(B) = 0.37 \times 0.53 + 0.44 \times 0.47\).
06

Substitute and Solve

Substitute the values into the equation: \(P(B) = 0.37 \times 0.53 + 0.44 \times 0.47 = 0.1961 + 0.2068 = 0.4029\). Now, find \(P(A|B)\): \(P(A|B) = \frac{0.37 \times 0.53}{0.4029} = \frac{0.1961}{0.4029} \approx 0.4864\).

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Key Concepts

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

Understanding Probability Basics
Probability is a way to measure how likely an event is to occur. Think of it as a scale from 0 to 1, where 0 means the event will not happen and 1 means the event will certainly occur. Here are some basic ideas to grasp:
  • If something is certain to happen, its probability is 1 (or 100%).
  • If something can never happen, its probability is 0 (or 0%).
  • A probability greater than 0 and less than 1 means there's a chance it could happen, but it's not guaranteed.
To calculate probability, you can use the formula: \[P(A) = \text{Number of favorable outcomes} / \text{Total number of possible outcomes}\]In the exercise, the probability that someone voted for Scott Walker was given as 0.53, or 53%. This means in a group of people, over half are likely to have voted for him.
Exploring Conditional Probability
Conditional probability narrows down the focus by looking at the chance of an event happening in the context of another event happening. It's like asking, "What are the chances of it raining today, given that the sky is already cloudy?" In our exercise, we use conditional probability to find the probability that someone voted for Scott Walker given that they have a college degree. The formula used is: \[P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}\]Here's what each part means:
  • \(P(A|B)\) is the probability of \(A\) happening, given that \(B\) has happened.
  • \(P(B|A)\) is the probability of \(B\) occurring if \(A\) happens, which was given as 0.37 in the exercise.
  • \(P(A)\) is the probability of \(A\) on its own, which was 0.53.
  • \(P(B)\) is the probability of \(B\) occurring overall, which we calculated using total probability.
Total Probability Theorem
The total probability theorem is a useful tool when you want to find the probability of an event that can happen in several ways. It says that you can calculate the overall probability of an event by adding up the probabilities of each different way it could happen, weighted by how likely each scenario is.In our exercise, we needed to find \(P(B)\), the probability that a randomly selected person from the exit poll has a college degree. To arrive at this, we considered:
  • \(P(B|A)\): the probability they have a degree, given they voted for Walker (0.37).
  • \(P(A)\): the probability that a person voted for Walker (0.53).
  • \(P(B|A^c)\): the probability they have a degree, given they voted against Walker (0.44).
  • \(P(A^c)\): the probability of voting against Walker (0.47).
Using the total probability theorem, we combined these to find:\[P(B) = P(B|A) \times P(A) + P(B|A^c) \times P(A^c)\]By calculating this, we achieve a full understanding of how often the event (having a college degree) occurs. In our example, this probability was approximately 0.4029, or 40.29%.

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Most popular questions from this chapter

Consider the following card game with a well-shuffled deck of cards. If you draw a red card, you win nothing. If you get a spade, you win \(\$ 5\). For any club, you win \(\$ 10\) plus an extra \(\$ 20\) for the ace of clubs. (a) Create a probability model for the amount you win at this game. Also, find the expected winnings for a single game and the standard deviation of the winnings. (b) What is the maximum amount you would be willing to pay to play this game? Explain your reasoning.

Ice cream usually comes in 1.5 quart boxes (48 fluid ounces), and ice cream scoops hold about 2 ounces. However, there is some variability in the amount of ice cream in a box as well as the amount of ice cream scooped out. We represent the amount of ice cream in the box as \(X\) and the amount scooped out as \(Y\). Suppose these random variables have the following means, standard deviations, and variances: $$ \begin{array}{cccc} \hline & \text { mean } & \text { SD } & \text { variance } \\ \hline X & 48 & 1 & 1 \\ Y & 2 & 0.25 & 0.0625 \\ \hline \end{array} $$ (a) An entire box of ice cream, plus 3 scoops from a second box is served at a party. How much ice cream do you expect to have been served at this party? What is the standard deviation of the amount of ice cream served? (b) How much ice cream would you expect to be left in the box after scooping out one scoop of ice cream? That is, find the expected value of \(X-Y\). What is the standard deviation of the amount left in the box? (c) Using the context of this exercise, explain why we add variances when we subtract one random variable from another.

In a new card game, you start with a well-shuffled full deck and draw 3 cards without replacement. If you draw 3 hearts, you win \(\$ 50\). If you draw 3 black cards, you win \(\$ 25\). For any other draws, you win nothing. (a) Create a probability model for the amount you win at this game, and find the expected winnings. Also compute the standard deviation of this distribution. (b) If the game costs \(\$ 5\) to play, what would be the expected value and standard deviation of the net profit (or loss)? (Hint: profit \(=\) winnings \(-\) cost \(; X-5\) ) (c) If the game costs \(\$ 5\) to play, should you play this game? Explain.

Backgammon. Backgammon is a board game for two players in which the playing pieces are moved according to the roll of two dice. Players win by removing all of their pieces from the board, so it is usually good to roll high numbers. You are playing backgammon with a friend and you roll two 6 s in your first roll and two 6 s in your second roll. Your friend rolls two 3 s in his first roll and again in his second row. Your friend claims that you are cheating, because rolling double 6 s twice in a row is very unlikely. Using probability, show that your rolls were just as likely as his.

Swaziland has the highest HIV prevalence in the world: \(25.9 \%\) of this country's population is infected with HIV. \(^{88}\) The ELISA test is one of the first and most accurate tests for HIV. For those who carry HIV, the ELISA test is \(99.7 \%\) accurate. For those who do not carry HIV, the test is \(92.6 \%\) accurate. If an individual from Swaziland has tested positive, what is the probability that he carries HIV?
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