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Chapter 4: Problem 7

Express all probabilities as fractions. In a horse race, a quinela bet is won if you selected the two horses that finish first and second, and they can be selected in any order. The 140 th running of the Kentucky Derby had a field of 19 horses. What is the probability of winning a quinela bet if random horse selections are made?

Short Answer

Expert verified
The probability of winning a quinela bet with random horse selections is \( \frac{1}{171} \).

Step by step solution

01

- Determine the total number of possible outcomes

Given that there are 19 horses in the race, first calculate the total number of possible outcomes for selecting 2 horses out of 19. This can be done using the combination formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] where \( n \) is the total number of items (horses) and \( k \) is the number of selections (2 in this case).Applying the values, we get:\[ \binom{19}{2} = \frac{19!}{2!(19-2)!} = \frac{19!}{2! \times 17!} = \frac{19 \times 18}{2 \times 1} = 171 \]So, there are 171 possible outcomes.
02

- Determine the number of successful outcomes

A successful outcome is the one where the two selected horses finish first and second in any order. There is only 1 successful combination of the two specific horses you picked out of the 171 possible combinations.
03

- Calculate the probability of a successful outcome

The probability \( P \) of winning a quinela bet can be calculated as the ratio of the number of successful outcomes to the total number of possible outcomes:\[ P(\text{win}) = \frac{\text{Number of successful outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{171} \]Thus, the probability of winning a quinela bet with random horse selections is \( \frac{1}{171} \).

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Key Concepts

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

combinatorial probability
Combinatorial probability involves calculating the likelihood of an event by considering all possible outcomes of a combination or arrangement of objects. For instance, if you want to know how many ways you can choose 2 horses out of 19 in a race, you use the combination formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]. Here, \( n = 19 \) and \( k = 2 \). This calculation tells you that there are 171 different ways to pick any 2 horses from the 19, which is crucial in determining probabilities in such scenarios. Understanding this formula is essential as it lays the foundation for the following calculations.
horse racing outcomes
In horse racing, outcomes refer to the different possible ways the horses can finish. If you bet on which horses will finish first and second, you are interested in 'quinela' bets. Each unique pairing of two horses, irrespective of their order, is an outcome. Suppose there are 19 horses in the race. If you pick any 2, the number of outcomes will be 171, considering all unique pairs. Recognizing all possible outcomes is vital for determining the probability of any single event such as winning a quinela bet.
quinela bet mathematics
A quinela bet in horse racing involves selecting two horses to finish first and second in any order. The math behind it can be intriguing yet straightforward. To calculate the probability of winning, you must first determine the total number of pairings using combinations. For 19 horses, you have \[ \binom{19}{2} = 171 \] possible pairings. You win if your selected pair comes in first and second in any order. Hence, for one specific pair, the probability is simply \[ \frac{1}{171} \], derived from the number of successful outcomes divided by the total possible outcomes.
probability fractions
Probability can often be expressed as fractions. For example, the chance of winning a quinela bet, given random horse selections, is calculated as the ratio of successful outcomes (selecting the correct 2 horses) to the total possible outcomes. In this scenario, the formula \[ P(\text{win}) = \frac{1}{171} \] represents the probability fraction. Ensure all probability fractions are in their simplest form to make them more interpretable and understandable.

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Most popular questions from this chapter

Express all probabilities as fractions. A clinical test on humans of a new drug is normally done in three phases. Phase I is conducted with a relatively small number of healthy volunteers. For example, a phase I test of bexarotene involved only 14 subjects. Assume that we want to treat 14 healthy humans with this new drug and we have 16 suitable volunteers available. a. If the subjects are selected and treated one at a time in sequence, how many different sequential arrangements are possible if 14 people are selected from the 16 that are available? b. If 14 subjects are selected from the 16 that are available, and the 14 selected subjects are all treated at the same time, how many different treatment groups are possible? c. If 14 subjects are randomly selected and treated at the same time, what is the probability of selecting the 14 youngest subjects?

Express all probabilities as fractions. As of this writing, the Powerball lottery is run in 44 states. Winning the jackpot requires that you select the correct five different numbers between 1 and 69 and, in a separate drawing, you must also select the correct single number between 1 and \(26 .\) Find the probability of winning the jackpot.

Acceptance Sampling. With one method of a procedure called acceptance sampling, a sample of items is randomly selected without replacement and the entire batch is accepted if every item in the sample is found to be okay. Involve acceptance sampling. The National Oceanic and Atmospheric Administration (NOAA) inspects seafood that is to be consumed. The inspection process involves selecting seafood samples from a larger "lot." Assume a lot contains 2875 seafood containers and 288 of these containers include seafood that does not meet inspection requirements. What is the probability that 3 selected container samples all meet requirements and the entire lot is accepted based on this sample? Does this probability seem adequate?

Express all probabilities as fractions. Credit card numbers typically have 16 digits, but not all of them are random. a. What is the probability of randomly generating 16 digits and getting your MasterCard number? b. Receipts often show the last four digits of a credit card number. If only those last four digits are known, what is the probability of randomly generating the other digits of your MasterCard number? c. Discover cards begin with the digits 6011 . If you know that the first four digits are 6011 and you also know the last four digits of a Discover card, what is the probability of randomly generating the other digits and getting all of them correct? Is this something to worry about?

Express the indicated degree of likelihood as a probability value between 0 and \(1 .\) When making a random guess for an answer to a multiple choice question on an SAT test, the possible answers are \(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}\), so there is 1 chance in 5 of being correct.
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