Category:
Mathematics

RESPOND TO DISCUSSION QUESTION AND RESPONSES ACCORDINGLY (160 WORDS EACH)

**Lotteries**

1. Use a state that has a lottery and calculate the number of possibilities for the numbers. Be sure to specify the maximum number (like 53) for the particular state’s lottery

2**. Given 32 flavors of ice cream, how many arrangements are possible on a double-scoop cone? How many arrangements are possible on a triple-scoop cone? What is the probability of getting a triple scoop of the same flavor? Explain the reasoning used to obtain the answers to the previous questions.**

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RESPOND ACCORDINGLY TO OTHER STUDENTS DISCUSSION RESPONSES TO THE QUESTION #2 ABOVE.

2DS. I found out that if you multiply the possibility of each scoop with one another, it works. So if the scoops must all be different flavors, then i would be 1/32 times 1 flavor out of the 31 possible flavors. (1/31)…On a triple scoop cone, it’d be 1/32 times 1/31 times 1/30. If the probability is going to be the same then its going to be 1/32, 1/32, 1/32, because repetition can happen, the scoops can ultimately be the same and there would be a total of 32 different flavors to choose from..

To break it down a little better…

Different: 1/32 x 1/31 x 1/30

Same: 1/32 x 1/32 x 1/32

2KM. For the double scoop cone different possible combos are = 32*32= 1024. This is the total possible combinations accounting for repeat flavors. The possible double scoop cone not allowing the same flavor for both scopes is 32*31 or 992. The triple scoop combo allowing for repeats is 32*32*32= 32768. Using reasoning I think the possibility of getting the same flavor on all three scopes on a random triple scoop cone is 32/32768. The denominator is the total possible combinations and the numerator represents the total ice cream flavor, there are only as many ways to make the same flavor as there is flavors.

2TH. If there are 32 flavors of ice cream and you order a double scoop, then there are 1024 possible arrangements to the cone.

32×32=1024

If there are 32 flavors of ice cream and you order a double scoop, then there is a much larger arrangement possibility….32768.

32x32x32=32768

As for the probability of getting a triple scoop of the same flavor you have to assume that the scoops are randomly chosen. There are 32 flavors so the likely hood of getting two scoops of the same flavor would be 1/32. The same would apply to the third scoop, the probability would be 1/32. Therefore, the probability that all three scoops would be the same flavor would be 1/1024 or 1/32*1/32 or another way to look at is to divide the total number of flavors, 32 by the total number of possible arrangements of a triple scoop, 32768. 32/32768 = 1/1024.

2SM. In probability questions such as these, you multiply the possiblity of each scoop with one another. If the scoops must all be different flavors, then it’s be 1 flavor out of 32 flavors or 1/32 times 1 flavor out of 31 flavors (because one is used) or 1/31. On a triple scoop cone, it’d be 1/32 times 1/31 times 1/30. If the probability is of getting a triple scoop of the same flavor it’d be 1/32 times 1/32 times 1/32 because repetition is possible, the scoops can be the same and there’d be 32 flavors to choose from

Different: 1/32 x 1/31 x 1/30

Same: 1/32 x 1/32 x 1/32

3. How might you construct a discrete probability distribution? Provide an example.

RESPOND ACCORDINGLY TO OTHER STUDENTS DISCUSSION RESPONSES TO THE QUESTION #3 ABOVE.

3DS. I would first realize the random variable that you are working with as discrete. Discrete variable has many different outcomes whereas a continuous doesn’t. For an example when counting the number of siblings you have that would be a discrete variable.

To construct a discrete variable you could draw a two sided table. The left said would be “x” to represent outcomes and then label the right side “p(x)” to represent probability of each outcome.

Write data down on the left side of the discrete probably distribution. So if you are given test scores you’d write each one of those scores down on the left side of the table. Each score would be in ascending order. To find the probability of each occurrence you would use the data that was given and then divide the total number of each outcome by the total amount of the sample (right side). Example 10 people out of 25 people get a score of “1” on the the test, divide 10 by 25 to get 40%. The number that you get is the probability that a member of the group will get a “1” on the the test. For this to work, each probability must equal a number between 0 and 1 and when you add all the probabilities together they all have to equal to one.

3SM. Identify the random variable you are working with as discrete or continuous. A discrete variable has a countable amount of outcomes whereas a continuous variable does not. For example, the amount of siblings you have is a discrete variable, because you can count the number of siblings you have. The precise height of a given person is a continuous variable, because when calculated to the decimal place, the number of decimals is infinite.

2. Draw a two-sided table. Label the left side of the table “x” to represent your possible outcomes (given in your data set) and the right side “P(x)” to represent the probability of each outcome.

3. Write your data set down in the left side of your discrete probability distribution. If you were given a set of scores, you would write each score in ascending order down the left side of your table. For example, if the possible scores ranged from 1 to 5, you would make five rows, labelled 1 through 5.

4. Find the probability of each occurrence. Using the data given to you divide the total number of each outcome by the total amount of sample. For example, if 20 people out of 50 got a score of “1” on a test, divide 20 by 50 to get 0.4, or 40 percent. The number you get is the probability that a member of the group will get a “1” on his test. Each probability must equal a number between zero and one. When added together, the sum of all of the probabilities must equal.

Source: http://www.ehow.com/how_8795663_might-construct-discrete-probability-distribution.html