/*! 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 124 A state runs a lottery in which ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 124

A state runs a lottery in which six numbers are randomly selected from \(40,\) without replacement. A player chooses six numbers before the state's sample is selected. (a) What is the probability that the six numbers chosen by a player match all six numbers in the state's sample? (b) What is the probability that five of the six numbers chosen by a player appear in the state's sample? (c) What is the probability that four of the six numbers chosen by a player appear in the state's sample? (d) If a player enters one lottery each week, what is the expected number of weeks until a player matches all six numbers in the state's sample?

Short Answer

Expert verified
(a) 1 in 3,838,380. (b) Approximately 5.31 x 10^(-5). (c) Approximately 0.002193. (d) 3,838,380 weeks.

Step by step solution

01

Total Possible Combinations

To solve the problem, we first calculate the total number of possible combinations of six numbers chosen from a set of 40 numbers. This is a combination problem, which can be calculated using the formula for combinations: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose. For this problem, \( n = 40 \) and \( r = 6 \), so the total number of combinations is \( \binom{40}{6} \). Compute: \[ \binom{40}{6} = \frac{40!}{6!(40-6)!} = 3,838,380 \]
02

Probability of Matching All Six Numbers

The probability is determined by the ratio of favorable outcomes to the total possible outcomes. Since there is exactly one way to match all six numbers, the probability \( P(A) \) is: \[ P(A) = \frac{1}{\binom{40}{6}} = \frac{1}{3,838,380} \]
03

Probability of Matching Five of Six Numbers

To find the probability of matching exactly five numbers, we first choose 5 correct numbers from the player's 6 numbers, which can be done in \( \binom{6}{5} = 6 \) ways. We also need to choose 1 incorrect number from the remaining 34 numbers. This can be done in \( \binom{34}{1} = 34 \) ways. Therefore, the favorable combinations are \( 6 \times 34 \). The probability \( P(B) \) is: \[ P(B) = \frac{6 \times 34}{\binom{40}{6}} = \frac{204}{3,838,380} = \frac{1}{18,826.37} \approx 5.31 \times 10^{-5} \]
04

Probability of Matching Four of Six Numbers

To find the probability of matching exactly four numbers, choose 4 correct numbers from the player's 6, which is \( \binom{6}{4} = 15 \). Then, choose 2 incorrect numbers from the 34 non-matching numbers, which is \( \binom{34}{2} = 561 \). Therefore, the favorable combinations are \( 15 \times 561 \). The probability \( P(C) \) is: \[ P(C) = \frac{15 \times 561}{\binom{40}{6}} = \frac{8,415}{3,838,380} = \frac{1}{456.01} \approx 0.002193 \]
05

Expected Number of Weeks to Win

The expected number of weeks until a player matches all six numbers is the reciprocal of the probability of matching all six numbers. From Step 2, we found \( P(A) = \frac{1}{3,838,380} \). Thus, the expected number of weeks \( E \) is: \[ E = \frac{1}{P(A)} = 3,838,380 \]

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.

Combinations
In probability, combinations are an essential concept when dealing with scenarios where the order of items does not matter. This is crucial for solving many probability problems, especially in games and lotteries. When you aim to choose a subset of a larger set without regard to the sequence of items, you are dealing with combinations.
For instance, in our lottery problem, we are interested in choosing 6 numbers from a total of 40. The formula for combinations is given by:
  • \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \)
where:
  • \( n \) is the total number of available items (in this case, 40 numbers)
  • \( r \) is the number of items you are choosing (for our problem, 6 numbers)
By applying this formula, you can determine the number of possible combinations, which is crucial for calculating the probability of various outcomes.
Lottery
The lottery is a classic example to illustrate the application of probability principles, especially combinations and calculations of likelihood for possible outcomes. In our scenario, a player chooses 6 numbers in hopes of matching them with the state's randomly selected subset of numbers.
Understanding this situation requires calculating the likelihood of different outcomes:
  • Matching all 6 numbers, which is the rarest event with only one favorable outcome.
  • Matching exactly 5 or 4 numbers, which are less rare but still unlikely events with more possible combinations.
The rarity of these events makes lotteries enticing and also underlines the skillful application of mathematical principles to analyze these probabilities.
Expected Value
Expected value is a powerful tool in probability, giving us insights into the average outcome of a random event if repeated many times. In the case of lottery games, it helps players understand the expected number of games or periods needed to achieve a specific outcome.
In our lottery problem, the expected value is used to determine how many draws it might take until a player matches all 6 numbers. The formula for expected value is simple:
  • Expected Value \( E = \frac{1}{P(A)} \)
Where \( P(A) \) is the probability of winning a single game. Knowing this helps players set expectations and gauge whether their strategies or expenditures align with their goals and dreams.
Probability of Events
Probability is the measure of the likelihood that an event will occur. In the context of lotteries, calculating probabilities involves figuring out how often certain outcomes are possible, relative to all possible outcomes. Probability can be formally expressed as:
  • \( P(Event) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} \)
In our example, events vary from matching all six lottery numbers to matching just four or five. Each has its own distinct probability:
  • Matching all 6 numbers: Extremely low probability, as shown in calculations.
  • Matching 5 or 4 numbers: Still very low, but relatively higher than six matches.
Understanding these probabilities helps illustrate just how challenging winning a lottery can be and underscores the importance of using probability to make informed predictions.

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

Suppose \(X\) has a hypergeometric distribution with \(N=100, n=4,\) and \(K=20 .\) Determine the following: (a) \(P(X=1)\) (b) \(P(X=6)\) (c) \(P(X=4)\) (d) Determine the mean and variance of \(X\).

Suppose \(X\) has a hypergeometric distribution with \(N=10, n=3,\) and \(K=4 .\) Sketch the probability mass function of \(X\). Determine the cumulative distribution function for \(X\)

Show that the probability density function of a negative binomial random variable equals the probability density function of a geometric random variable when \(r=1\). Show that the formulas for the mean and variance of a negative binomial random variable equal the corresponding results for a geometric random variable when \(r=1\).

The number of telephone calls that arrive at a phone exchange is often modeled as a Poisson random variable. Assume that on the average there are 10 calls per hour. (a) What is the probability that there are exactly five calls in one hour? (b) What is the probability that there are three or fewer calls in one hour? (c) What is the probability that there are exactly 15 calls in two hours? (d) What is the probability that there are exactly five calls in 30 minutes?

Consider a sequence of independent Bernoulli trials with \(p=0.2\). (a) What is the expected number of trials to obtain the first Success? (b) After the eighth success occurs, what is the expected number of trials to obtain the ninth success?
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

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.