/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Candy Someone hands you a box of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Candy Someone hands you a box of a dozen chocolatecovered candies, telling you that half are vanilla creams and the other half peanut butter. You pick candies at random and discover the first three you eat are all vanilla. a. If there really were 6 vanilla and 6 peanut butter candies in the box, what is the probability that you would have picked three vanillas in a row? b. Do you think there really might have been 6 of each? Explain. c. Would you continue to believe that half are vanilla if the fourth one you try is also vanilla? Explain.

Short Answer

Expert verified
The probability is about 0.091 or 9 in 100 repeats of the experiment. Drawing three or even four vanilla candies in a row, doesn't necessarily imply that there are more than 6 vanilla candies in the box, but we might start to question that assumption if we continue to draw more vanilla candies.

Step by step solution

01

Calculate the probability for a single draw

Initially, there are 6 vanilla and 6 peanut butter candies, giving us a total of 12 candies. For the first draw, the probability of getting a vanilla candy is therefore \(\frac{6}{12} = 0.5\).
02

Calculate the cumulative probability

To find the probability of picking vanilla on the three draws in a row, it's necessary to multiply the probabilities of each draw. After the first candy is drawn, there are only 11 candies left, with 5 of them being vanilla. The probability of the second candy being vanilla is \(\frac{5}{11}\). The third draw has now a total of 10 candies with 4 of them being vanilla, which gives us a probability of \(\frac{4}{10}\). The cumulative probability for the three draws hence is \(\frac{6}{12} * \(\frac{5}{11} * \(\frac{4}{10} = 0.091.\)
03

Interpret the result

After calculating the probability, it's important to understand its implications. The calculated probability of 0.091 indicates that, on average, the scenario of drawing three vanillas in a row would happen about 9 out of 100 times when pulling randomly from the box. Thus, just because we drew three vanillas right now, we shouldn't assume that there are more than 6 vanilla candies. As for drawing a fourth vanilla, it can happen but it is unlikely as the probability is \(\frac{4}{10}\) or 0.4, and drawing a fourth one wouldn't necessarily mean that there are more vanilla candies, but we may start questioning the assumption.

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.

Cumulative Probability
Cumulative probability refers to the likelihood of an event or series of events occurring sequentially in specific conditions. It's akin to a chain, where each link represents an individual event and the strength of the chain (the cumulative probability) is determined by the combined strength of all links. In our exercise involving the selection of vanilla candies, calculating the cumulative probability required considering the decreasing number of candies after each draw.

It's essential to view this as a multi-stage process. After enjoying the first candy, the box now contains 11 candies with 5 vanillas. The probability for this next stage is then recalculated based on the new numbers. This step-wise reduction continues, affecting the probability of each subsequent event. Understanding this concept is crucial in a plethora of fields, from quality control to decision making under uncertainty. The principle has wide-reaching implications, whether it's forecasting weather events or analyzing the likelihood of a medical treatment leading to consecutive successful outcomes.
Probability Calculations
Probability calculations are fundamental tools used to quantify the likelihood of an event. These calculations can become complex when dealing with multiple events, especially in conditional probabilities or when the outcome of one event affects another. In our sweet dilemma with the candies, the initial probability for picking a vanilla was simple: \(\frac{6}{12} = 0.5\). However, we needed to adjust our calculations as each candy was selected.

The key to undertaking probability calculations is recognizing the evolving nature of the scenario – each draw changed the total number of candies and the vanilla count. By systematically adjusting our probability calculations after each draw, we arrived at a composite or cumulative probability that reflected the likelihood of selecting three vanillas consecutively. Performing these layered calculations, one must pay careful attention to each step to ensure an accurate overall assessment of probabilities.
Probability Interpretation
Interpreting the results of probability calculations is as important as the calculations themselves. While the numbers provide raw data, interpreting them gives us insight into the context and potential implications. From the exercise, we determined a cumulative probability of 0.091, or about 9% chance, of picking three vanilla candies consecutively. This interpretation suggests that while it is possible to pick three vanillas in a row, it is a relatively unlikely event.

The interpretation can influence our belief in the initial assumption that there are an equal number of vanilla and peanut butter candies. If such a scenario seemed more likely than calculated (for example, if we pulled three vanillas out of the box several times in a row), one might suspect that the box might actually contain more than six vanilla candies. However, the occurrence of one unlikely event does not nullify the original assumption; other supporting evidence would be needed to form a more definitive judgment. The interpretation of probability merges mathematical results with critical thinking, guiding our decisions and expectations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Hypotheses and parameters As in Exercise 3 ?, for each of the following situations, define the parameter and write the null and alternative hypotheses in terms of parameter values. a. Seat-belt compliance in Massachusetts was \(65 \%\) in 2008. The state wants to know if it has changed. b. Last year, a survey found that \(45 \%\) of the employees were willing to pay for on-site day care. The company wants to know if that has changed. c. Regular card customers have a default rate of \(6.7 \% .\) A credit card bank wants to know if that rate is different for their Gold card customers. d. Regular card customers have been with the company for an average of 17.3 months. The credit card bank wants to know if their Gold card customers have been with the company on average the same amount of time.

GRE performance again Instead of advertising the percentage of customers who improve by at least 10 points, a manager suggests testing whether the mean score improves at all. For each customer they record the difference in score before and after taking the course (After \(-\) Before). a. State the null and alternative hypotheses. b. The P-value from the test is 0.65 . Does this provide any evidence that their course works? c. From part \(b,\) what can you tell, if anything, about the mean difference in the sample scores?

Dice The seller of a loaded die claims that it will favor the outcome \(6 .\) We don't believe that claim, and roll the die 200 times to test an appropriate hypothesis. Our P-value turns out to be 0.03. Which conclusion is appropriate? Explain. a. There's a \(3 \%\) chance that the die is fair. \(\mathrm{b}\). There's a \(97 \%\) chance that the die is fair. c. There's a \(3 \%\) chance that a loaded die could randomly produce the results we observed, so it's reasonable to conclude that the die is fair. d. There's a \(3 \%\) chance that a fair die could randomly produce the results we observed, so it's reasonable to conclude that the die is loaded.

Take the offer II We saw in Chapter 16 ?, Exercise 36 ? that First USA tested the effectiveness of a double miles campaign by recently sending out offers to a random sample of 50,000 cardholders. Of those, 1184 registered for the promotion. Even though this is nearly a \(2.4 \%\) rate, a staff member suspects that the success rate for the full campaign will be no different than the standard \(2 \%\) rate that they are used to seeing in similar campaigns. What do you predict? a. What are the hypotheses? b. Are the assumptions and conditions for inference met? c. Do you think the rate would change if they use this fundraising campaign? Explain.

Absentees The National Center for Education Statistics monitors many aspects of elementary and secondary education nationwide. Their 1996 numbers are often used as a baseline to assess changes. In \(1996,34 \%\) of students had not been absent from school even once during the previous month. In a 2000 survey, responses from 8302 students showed that this figure had slipped to \(33 \%\). Officials would, of course, be concerned if student attendance were declining. Do these figures give evidence of a change in student attendance? a. Write appropriate hypotheses. b. Check the assumptions and conditions. c. Perform the test and find the P-value. d. State your conclusion. e. Do you think this difference is meaningful? Explain.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.