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Big loser in Lotto Example 10 showed that the probability of having the winning ticket in Lotto South was 0.00000007 . Find the probability of holding a ticket that has zero winning numbers out of the 6 numbers selected (without replacement) for the winning ticket out of the 49 possible numbers.

Short Answer

Expert verified
The probability is approximately 0.3174.

Step by step solution

01

Calculate the Total Number of Ways to Choose 6 Numbers from 49

We need to calculate the number of possible combinations of selecting 6 numbers out of 49. This is done using the combination formula: \( C(n, r) = \frac{n!}{r!(n-r)!} \). For our case, \( n = 49 \) and \( r = 6 \):\[ C(49, 6) = \frac{49!}{6!(49-6)!} = \frac{49!}{6! \times 43!}. \] Calculating this gives a total of 13,983,816 possible combinations.
02

Calculate the Total Number of Ways to Choose 6 Numbers from the 43 Non-Winning Numbers

The number of ways to select 6 numbers from the 43 non-winning numbers is calculated similarly:\[ C(43, 6) = \frac{43!}{6!(43-6)!} = \frac{43!}{6! \times 37!}. \]Calculating this results in 4,436,388 possible combinations of non-winning numbers.
03

Find the Probability of Holding a Ticket with Zero Winning Numbers

The probability of choosing a ticket with zero winning numbers is the ratio of choosing 6 non-winning numbers to the total number of combinations.So, the probability is:\[ \frac{C(43, 6)}{C(49, 6)} = \frac{4,436,388}{13,983,816}. \]This simplifies approximately to 0.3174.

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Key Concepts

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

Combination Formula
When tackling probability problems, especially in lottery games like Lotto South, understanding the combination formula is crucial. The combination formula helps determine how many ways you can choose a specific number of items from a larger set, without caring about the order of selection.

The formula is expressed as:
  • \( C(n, r) = \frac{n!}{r!(n-r)!} \)
Here, \( n \) stands for the total number of items, and \( r \) is the number of items to be chosen. The "!" symbol denotes a factorial, which implies multiplying a series of descending natural numbers. For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 \).

In the context of our lottery problem, \( n \) is 49 because there are 49 numbers to potentially choose from, and \( r \) is 6, representing the 6 numbers on a lottery ticket. By applying the combination formula, we can calculate the total number of possible 6-number combinations from 49 numbers, which is 13,983,816. This provides a foundation to further evaluate probabilities.
Probability Calculation
Once you know how many ways you can choose your numbers, calculating probability is the next step. Probability is all about measuring the chance of a particular outcome out of all possible outcomes.

In our exercise, we want to determine the likelihood of picking a ticket with zero winning numbers. This is essentially finding out how probable it is to select any combinations of numbers except the winning ones.
To find this probability:
  • Calculate the number of ways to pick 6 numbers from the 43 non-winning numbers using the combination formula. This is denoted as \( C(43, 6) \) which equals 4,436,388.
  • Then, divide this by the total number of combinations from 49 numbers, \( C(49, 6) = 13,983,816 \).
  • The probability formula simplifies to \( \frac{C(43, 6)}{C(49, 6)} \).
This division gives a result of approximately 0.3174, indicating that there is a 31.74% chance of holding a ticket with no winning numbers.
Non-Replacement Events
Understanding non-replacement events is another key element when calculating lottery probabilities. Non-replacement means once a number is drawn, it cannot be selected again. This affects probability calculations because the pool of possible numbers decreases with each selection.

In our scenario, when the first winning number is selected from 49 total numbers, it cannot be picked again, leaving 48 numbers for the next draw, and so on. This concept is crucial when calculating probabilities as it reflects real-world lottery rules where each number is unique and not replacable.

The non-replacement principle ensures that, for your chosen numbers (the ticket), you only select from those numbers not previously drawn as wins. This affects the number of non-winning or losing numbers. The importance of non-replacement is why when we calculate \( C(43, 6) \), we choose from a reduced pool of only 43 numbers. In contrast, when initially calculating \( C(49, 6) \), we used the entire pool, showing how probability shifts with non-replacement over sequential selections. This method models the true dynamics of lottery draws comprehensively.

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Most popular questions from this chapter

Before the first human heart transplant, Dr. Christiaan Barnard of South Africa was asked to assess the probability that the operation would be successful. Did he need to rely on the relative frequency definition or the subjective definition of probability? Explain.

Go Celtics! Larry Bird, who played pro basketball for the Boston Celtics, was known for being a good shooter. In games during \(1980-1982,\) when he missed his first free throw, 48 out of 53 times he made the second one, and when he made his first free throw, 251 out of 285 times he made the second one. a. Form a contingency table that cross tabulates the outcome of the first free throw (made or missed) in the rows and the outcome of the second free throw (made or missed) in the columns. b. For a given pair of free throws, estimate the probability that Bird (i) made the first free throw and (ii) made the second free throw. (Hint: Use counts in the (i) row margin and (ii) column margin.) c. Estimate the probability that Bird made the second free throw, given that he made the first one. Does it seem as if his success on the second shot depends strongly, or hardly at all, on whether he made the first?

Your friend is interested in estimating the proportion of people who would vote for his project in a local contest. He selects a large sample among his many friends and claims that, with such a large sample, he does not need to worry about the method of selecting the sample. What is wrong in this reasoning? Explain.

Explain what is meant by the long-run relative frequency definition of probability.

The powerrank.com website (http:// thepowerrank.com/2014/06/06/world- cup-2014-winprobabilities-from-the-power-rank/) listed the probability of each team to win the 2014 World Cup in soccer as follows: 1\. Brazil, \(35.9 \%\). 2\. Argentina, \(10.0 \%\). 3\. Spain, \(8.9 \%\). 4\. Germany, \(7.4 \%\). 5\. Netherlands, \(5.7 \%\). 6\. Portugal, \(3.9 \%\). 7\. France, \(3.4 \%\). 8\. England, \(2.8 \%\). 9\. Uruguay, \(2.5 \%\). 10\. Mexico, \(2.5 \%\). 11\. Italy, \(2.3 \%\). 12\. Ivory Coast, \(2.0 \%\), 13\. Colombia, \(1.5 \%\). 14\. Russia, \(1.5 \%\). 15\. United States, \(1.1 \%\). 16\. Chile, \(1.0 \%\). 17\. Croatia, \(0.9 \%\) 18\. Ecuador, \(0.8 \%\). 19\. Nigeria, \(0.8 \%\). 20\. Switzerland, \(0.7 \%\). 21\. Greece, \(0.6 \%\) 22\. \(\operatorname{Iran}, 0.6 \%\). 23\. Japan, \(0.6 \%\). 24\. Ghana, \(0.6 \%\). 25\. Belgium, \(0.4 \%\). 26\. Honduras, \(0.3 \%\). 27\. South Korea, \(0.3 \%\). 28\. Bosnia-Herzegovina, \(0.3 \%\). 29\. Costa Rica, \(0.3 \%\). 30\. Cameroon, \(0.2 \%\). 31\. Australia, \(0.2 \%\). 32\. Algeria, \(0.1 \%\). a. Interpret Brazil's probability of \(35.9 \%,\) which was based on computer simulations of the tournament. Is it a relative frequency or a subjective interpretation of probability? b. Germany would emerge as the actual winner of the 2014 World Cup. Does this indicate that the \(7.4 \%\) chance of Germany winning, which was calculated before the tournament, should have been \(100 \%\) instead?

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