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lutik1710
7 days ago
14

Five thousand tickets are sold at​ $1 each for a charity raffle. Tickets are to be drawn at random and monetary prizes awarded a

s​ follows: 1 prize of ​$500​, 3 prizes of ​$300​, 5 prizes of ​$50​, and 20 prizes of​ $5. What is the expected value of this raffle if you buy 1​ ticket?
Mathematics
1 answer:
babunello [8.4K]7 days ago
7 0

Response:

the expected value of this raffle if you purchase 1 ticket = -0.65

Breakdown of the calculation:

Details:

5,000 tickets are sold at​ $1 each for a charitable raffle

Winners will be chosen at random with cash prizes as follows: 1 prize of ​$500​, 3 prizes of ​$300​, 5 prizes of ​$50​, and 20 prizes of​ $5.

Therefore, the value and its respective probability can be calculated as follows:

Value                              Probability

$500 - $1 = $499              1/5000

$300 - $1 = $299              3/5000

$50 - $1 = $49                    5/5000

$5 - $1 = $4                     20/5000

-$1                           1 - 29/5000 = 4971/5000

The expected value of the raffle when buying 1 ticket is computed as follows:

E(x) = \sum x * P(x)

E(x) = (499 * \dfrac{1}{5000} + 299 *\dfrac{3}{5000} + 49 *\dfrac{5}{5000} + 4 * \dfrac{20}{5000} + (-1 * \dfrac{4971}{5000} ))

E(x) = (0.0998 + 0.1794+0.049 + 0.016 + (-0.9942 ))

E(x) = (0.3442 -0.9942 )

\mathbf{E(x) = -0.65}

So, the expected value of this raffle when one ticket is purchased = -0.65

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There are several possible outcomes. The initial composition of the urns is as follows: Urn 1 contains 2 red chips and 4 white chips, totaling 6 chips, whereas Urn 2 has 3 red and 1 white, amounting to 4 chips. When a chip is drawn from the first urn, the probabilities are as follows: for a red chip, it is probability is (2 red from 6 chips = 2/6 = 1/2); for a white chip, it is (4 white from 6 chips = 4/6 = 2/3). After the chip is transferred to the second urn, two scenarios can arise: if the chip drawn from the first urn is white, then Urn 2 will contain 3 red and 2 white chips, making a total of 5 chips, creating a 40% chance for drawing a white chip. Conversely, if a red chip is drawn first, Urn 2 will contain 4 red and 1 white chip, which results in a 20% chance of drawing a white chip. This scenario exemplifies a dependent event, as the outcome hinges on the type of chip drawn first from Urn 1. For the first scenario, the combined probability is (the probability of drawing a white chip from Urn 1) multiplied by (the probability of drawing a white chip from Urn 2), equaling 26.66%. For the second scenario, the probabilities yield a value of 6%.
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12 days ago
To better understand how husbands and wives feel about their finances, Money Magazine conducted a national poll of 1010 married
Svet_ta [9518]

Answer:

  • a. Refer to the table below
  • b. Refer to the table below
  • c. 0.548
  • d. 0.576
  • e. 0.534
  • f) i) 0.201, ii) 0.208

Explanation:

To begin with, organize the data provided:

Table: "Who excels at obtaining deals?"

                       Who Excels?

Respondent      I Am        My Spouse     We are Equal

Husband           278             127                 102

Wife                   290            111                   102

a. Create a joint probability table and utilize it to respond to the ensuing inquiries.

The joint probability table presents identical details expressed as proportions. The values from the table need to be divided by the total number of responses involved.

1. Total responses: 278 + 127 + 102 + 290 + 111 + 102 = 1,010.

2. Determine each proportion:

  • 278/1,010 = 0.275
  • 127/1,010 = 0.126
  • 102/1,010 = 0.101
  • 290/1,010 = 0.287
  • 111/1,010 = 0.110
  • 102/1,010 = 0.101

3. Construct the table containing these values:

Joint probability table:

Respondent      I Am        My Spouse     We Are Equal

Husband           0.275           0.126                 0.101

Wife                   0.287           0.110                  0.101

This table illustrates that the joint probability of identifying as a husband while choosing 'I am' equals 0.275. Each cell conveys the joint probability associated with each gender's response.

Consequently, this delineates the purpose of a joint probability table.

b. Generate marginal probabilities for Who Excels (I Am, My Spouse, We Are Equal). Provide commentary.

Marginal probabilities are computed for each row and column of the table, indicated in the margins, which is their namesake.

For the column titled "I am," it amounts to: 0.275 + 0.287 = 0.562

Similarly, perform calculations for the other two columns.

For the row designated 'Husband,' it would thus be 0.275 + 0.126 + 0.101 = 0.502. Apply the same for the row labeled 'Wife.'

Table Marginal probabilities:

Respondent      I Am        My Spouse     We Are Equal     Total

Husband           0.275           0.126                 0.101             0.502

Wife                   0.287           0.110              0.101             0.498

Total                 0.562           0.236            0.202             1.000

Notably, when summing the marginal probabilities for both rows and columns, the results will always equate to 1. This is a consistent truth for marginal probabilities.

c. Given the respondent is a husband, what is the likelihood that he believes he is better at securing deals than his wife?

This requires the utilization of conditional probability.

The goal here is to ascertain the probability of the response being "I am" when the respondent identifies as a "Husband."

Using conditional probability:

  • P ( "I am" / "Husband") = P ("I am" ∩ "Husband) / P("Husband")

  • P ("I am" ∩ "Husband) = 0.275 (obtained from the intersection of columns "I am" and rows "Husband")

  • P("Husband") = 0.502 (derived from total of row "Husband")

  • P ("I am" ∩ "Husband) / P("Husband") = 0.275 / 0.502 = 0.548

d. In the instance that the respondent is a wife, what probability exists that she believes she is superior to her husband in acquiring deals?

We seek to identify the probability wherein the response claims "I am" while the respondent is labeled a "Wife," applying the conditional probability formula again:

  • P ("I am" / "Wife") = P ("I am" ∩ "Wife") / P ("Wife")

  • P ("I am" / "Wife") = 0.287 / 0.498

  • P ("I am" / "Wife") = 0.576

e. When responding that "My spouse" is better at scoring deals, what is the likelihood that the claim originated from a husband?

We aim to compute: P ("Husband" / "My spouse")

Applying the conditional probability formula:

  • P("Husband" / "My spouse") = P("Husband" ∩ "My spouse")/P("My spouse")

  • P("Husband" / "My spouse") = 0.126/0.236

  • P("Husband" / "My spouse") = 0.534

f. When the response indicates "We are equal," what likelihood exists that this response is from a husband? What is the chance that it hails from a wife?

What is the likelihood that this response came from a husband?

  • P("Husband" / "We are equal") = P("Husband" ∩ "We are equal") / P ("We are equal")

  • P("Husband" / "We are equal") = 0.101 / 0.502 = 0.201

What is the chance the response originated from a wife:

  • P("Wife") / "We are equal") = P("Wife" ∩ "We are equal") / P("We are equal")

  • P("Wife") / "We are equal") = 0.101 / 0.498 = 0.208
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1 month ago
A statistician at a metal manufacturing plant is sampling the thickness of metal plates. If an outlier occurs within a particula
babunello [8423]

Answer: 26.3 mm

Step-by-step explanation:

The two-standard deviations rule for outliers states that any value lying outside two standard deviations from the average is considered an outlier.

Given that the Mean is 23.5 millimeters (mm) and the standard deviation is 1.4 mm

The maximum thickness that should be reviewed = mean + 2 (standard deviation)

= 23.5 + 2(1.4)

= 23.5 + 2.8

= 26.3 mm

therefore, the maximum thickness warranting machine configuration review by the statistician = 26.3 mm.

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To calculate, simply multiply 12 by 0.15 or 15%, which equals 1.8. Then subtract 1.8 from 12, yielding $10.20 as the discounted price for the pizza.

I hope this is helpful


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