/*! 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 44 Example 10 showed that the proba... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 44

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 of having zero winning numbers is 37.5%.

Step by step solution

01

Understand the Total Outcomes

The exercise involves selecting 6 numbers from 49. The total number of ways to choose 6 numbers from 49 is given by the combination formula \( \binom{n}{k} \). This represents all possible outcomes for selecting any 6 numbers from 49.
02

Calculate Total Outcomes

Use the combination formula \( \binom{49}{6} \) to calculate the total number of different combinations you can make with 6 numbers out of 49. The formula is \( \binom{49}{6} = \frac{49 \times 48 \times 47 \times 46 \times 45 \times 44}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 13,983,816 \).
03

Understand 'Zero Winning Numbers' Concept

To have zero winning numbers, none of the 6 chosen numbers must match any of the winning 6 numbers. Therefore, you can only choose from the remaining 43 numbers that were not selected in the winning ticket.
04

Calculate Favorable Outcomes for Zero Winning Numbers

We need to calculate the number of ways to choose 6 numbers from the 43 that were not winning numbers. Use the combination formula \( \binom{43}{6} \). This gives \( \binom{43}{6} = \frac{43 \times 42 \times 41 \times 40 \times 39 \times 38}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 5,245,786 \).
05

Determine the Probability of Zero Winning Numbers

The probability is calculated by taking the ratio of favorable outcomes to total outcomes. This is \( \frac{\binom{43}{6}}{\binom{49}{6}} = \frac{5,245,786}{13,983,816} \).
06

Simplify the Probability Fraction

Divide \(5,245,786\) and \(13,983,816\) by their greatest common divisor, resulting in \( \frac{5,245,786}{13,983,816} \approx 0.375 \) or 37.5%.

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 91影视!

Key Concepts

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

Combinatorics
Combinatorics is a fundamental aspect of probability theory that deals with counting and arranging objects in specific ways. Imagine you are trying to figure out how many unique groups of friends you can invite to your party from a larger group. This is where combinatorics comes in handy.
In problems like lotteries,
  • we need to select a certain number of items (like lottery numbers) from a larger set,
  • and the order in which we select them does not matter.
The tools you use in combinatorics, like the combination formula, enable you to calculate the number of possible combinations efficiently. This ensures you understand the scale and probability of different outcomes occurring.
Lottery Probability
Lottery probability involves calculating the chances of winning or achieving a specific result in a lottery game. In our case, we're calculating the chance of having a ticket with zero winning numbers.
The main points to understand include:
  • the total number of ways to select numbers (known as combinations),
  • and the number of ways that specifically meet your criteria (like having no winning numbers).
By comparing the favorable outcomes to the total possible outcomes, you can determine the probability. In this case, the mathematical approach reveals that despite the huge number of possible outcomes, the chance of getting no matches is significantly larger than winning.
Combination Formula
The combination formula is a crucial mathematical tool for solving lottery problems. It helps determine the number of ways to choose a subset of items from a larger set, where the order doesn鈥檛 matter.
Mathematically, it is expressed as:\[\binom{n}{k}=\frac{n!}{k!(n-k)!}\]
Here, 'n' is the total number of items to choose from, and 'k' is the number of items to select.
For example:
  • To find the probability of choosing 6 numbers from 49, use \(\binom{49}{6}\).
  • For our specific problem of no matches, we use \(\binom{43}{6}\), since we are picking from the numbers that weren鈥檛 winning numbers.
This formula is essential for simplifying complex combinations into simpler calculations.
Mathematical Calculations
Mathematical calculations are the backbone of solving lottery probability problems. They involve using accurate and careful steps to reach the right conclusion.
Begin with:
  • calculating total possible outcomes,
  • then determining favorable outcomes based on the problem's restriction.
The final step is to compute the probability by dividing the number of favorable outcomes by the total possible outcomes.
This is done as follows:\[\text{Probability} = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}\]
In our lottery exercise, we computed a probability of 37.5% for having zero winning numbers by dividing the combinations of non-winning numbers by the total combinations. This highlights the importance of understanding each step to ensure your calculations are both correct and meaningful.

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

A 2007 study by the National Center on Addiction and Substance Abuse at Columbia University reported that for college students, the estimated probability of being a binge drinker was 0.50 for males and 0.34 for females. Using notation, express each of these as a conditional probability.

A couple plans to have two children. Each child is equally likely to be a girl or boy, with gender independent of that of the other child. a. Construct a sample space for the genders of the two children. b. Find the probability that both children are girls. c. Answer part b if in reality, for a given child, the chance of a girl is 0.49 .

A pollster wants to estimate the proportion of Canadian adults who support the prime minister's performance on the job. He comments that by the law of large numbers, to ensure a sample survey's accuracy, he does not need to worry about the method for selecting the sample, only that the sample has a very large sample size. Do you agree with the pollster's comment? Explain.

Is there intelligent life on other planets in the universe? If you are asked to state the probability that there is, would you need to rely on the relative frequency or the subjective definition of probability? Explain.

Two friends decide to go to the track and place some bets. One friend remarks that in an upcoming race, the number 5 horse is paying 50 to \(1 .\) This means that anyone who bets on the 5 horse receives \(\$ 50\) for each \(\$ 1\) bet, if in fact the 5 horse wins the race. He goes on to mention that it is a great bet, because there are only eight horses running in the race, and therefore the probability of horse 5 winning must be \(1 / 8 .\) Is the last statement true or false? Explain.
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation
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.