/*! 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} Q2CRE South Carolina Pick 3 In South C... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

South Carolina Pick 3 In South Carolina’s Pick 3 lottery game, you can pay \(1 to select a sequence of three digits, such as 227. If you buy only one ticket and win, your prize is \)500 and your net gain is $499.

a. If you buy one ticket, what is the probability of winning?

b. If you play this game once every day, find the mean number of wins in years with exactly 365 days.

c. If you play this game once every day, find the probability of winning exactly once in 365 days.

d. Find the expected value for the purchase of one ticket.

Short Answer

Expert verified

a. The probability of winning is equal to 0.001.

b. The mean number of wins in a year with 365 days is equal to 0.365.

c. The probability of winning exactly once in a year is equal to 0.253.

d. The expected value for the purchase of one ticket is equal to – 50 cents.

Step by step solution

01

Given information

The number of moons of all the 8 planets is listed.

02

Probability of Winning

a.

The total number of possible sequences is computed below:

\(10 \times 10 \times 10 = 1000\)

The number of winning sequences = 1

The probability of winning is computed below:

\(\begin{array}{c}P\left( {{\rm{winning}}} \right) = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{winning}}\;{\rm{sequences}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{sequences}}}}\\ = \frac{1}{{1000}}\\ = 0.001\end{array}\)

Therefore, the probability of winning is equal to 0.001.

03

Mean

b.

The number of days in a year is equal to n=365.

The probability of winning a game is equal to p=0.001.

The mean number of wins in a year if one game is played every day is computed below:

\(\begin{array}{c}\mu = np\\ = 365\left( {0.001} \right)\\ = 0.365\end{array}\)

Thus, the mean number of wins in a year if one game is played every day is equal to 0.365.

04

Probability of winning once in a year

c.

Let X denote the number of wins in a year.

So, X follows a Poisson distribution with mean equal to 0.365.

The probability of winning once in a year is computed below:

\[\begin{array}{c}P\left( {X = x} \right) = \frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}\\P\left( {X = 1} \right) = \frac{{{{\left( {0.365} \right)}^1}{{\left( {2.71828} \right)}^{ - 0.365}}}}{{1!}}\\ = 0.253\end{array}\]\(\left( {{x_{{\rm{loss}}}}} \right)\)

Thus, the probability of winning once in a year is equal to 0.253.

05

Expected value

d.

The amount received on winning\(\left( {{x_{{\rm{win}}}}} \right)\)is equal to $499.

The probability of winning\(\left( {{p_{win}}} \right)\)is equal to 0.001.

The amount lost on losing is equal to -1dollar.

The probability of losing is equal to:

\(\begin{array}{c}{p_{{\rm{loss}}}} = 1 - {p_{win}}\\ = 1 - 0.001\\ = 0.999\end{array}\)

The expected value lost/gained on the purchase of one ticket is computed below:

\(\begin{array}{c}E = {x_{win}}{p_{win}} + {x_{loss}}{p_{loss}}\\ = \left( {499} \right)\left( {0.001} \right) + \left( { - 1} \right)\left( {0.999} \right)\\ = - 0.50\end{array}\)

Thus, the expected value for the purchase of one ticket is equal to – 0.50 dollars or – 50 cents.

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!

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

Significance with Range Rule of Thumb. In Exercises 31 and 32, assume that hybridization experiments are conducted with peas having the property that for offspring, there is a 0.75 probability that a pea has green pods (as in one of Mendel’s famous experiments).

Hybrids Assume that offspring peas are randomly selected in groups of 16.

a. Find the mean and standard deviation for the numbers of peas with green pods in the groups of 16.

b. Use the range rule of thumb to find the values separating results that are significantly low or significantly high.

c. Is a result of 7 peas with green pods a result that is significantly low? Why or why not?

In Exercises 1–5, assume that 74% of randomly selected adults have a credit card (basedon results from an AARP Bulletin survey). Assume that a group of five adults is randomlyselected.

Find the probability that at least one of the five adults has a credit card. Does the result apply to five adult friends who are vacationing together? Why or why not?

Geometric Distribution If a procedure meets all the conditions of a binomial distribution except that the number of trials is not fixed, then the geometric distribution can be used. The probability of getting the first success on the xth trial is given by , where p is the probability of success on any one trial. Subjects are randomly selected for the National Health and Nutrition Examination Survey conducted by the National Center for Health Statistics, Centers for Disease Control and Prevention. The probability that someone is a universal donor (with group O and type Rh negative blood) is 0.06. Find the probability that the first subject to be a universal blood donor is the fifth person selected.

In Exercises 15–20, assume that random guesses are made for eight multiple choice questions on an SAT test, so that there are n = 8 trials, each with probability of success (correct) given by p = 0.20. Find the indicated probability for the number of correct answers.

Find the probability that the number x of correct answers is at least 4.

In Exercises 6–10, use the following: Five American Airlines flights are randomly selected, and the table in the margin lists the probabilities for the number that arrive on time (based on data from the Department of Transportation). Assume that five flights are randomly selected.

Does the table describe a probability distribution?

x

P(x)

0

0+

1

0.006

2

0.051

3

0.205

4

0.409

5

0.328

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.