Macy rolls a fair eight sided die with sides numbered from 1 to 8. Then from a box, she draws a numbered slip of paper. The slips of paper have numbers shown below.
1,2,5,7,8
Finally, she calculates the sum of her results.
Select all the true statements.
1.)A sum equal to 8 is the result 20 times in 80 rounds. This suggests the game is unfair.
2.)The probability of getting a sum that is a multiple of 3 is 3/8
3.) The probability of getting a sum that is even is 1/2.
4.) The probability of getting a sum that is greater than or equal to 12 is 11/40.
5.) The probability of getting a sum that is less than 10 is 21/40
There can be multiple answers

Probabilities are used to determine the chances of an event. The true statements from the 5 options are:

Given that:

[tex]Die =\{1,2,3,4,5,6,7,8\}[/tex]

[tex]Number = \{1,2,5,7,8\}[/tex]

The possible outcomes are:[tex]S = \{(1,1),(1,2),(1,5),(1,7),(1,8),(2,1),(2,2),(2,5),(2,7),(2,8),(3,1),(3,2),(3,5),(3,7),(3,8)\\(4,1),(4,2),(4,5),(4,7),(4,8),(5,1),(5,2),(5,5),(5,7),(5,8),(6,1),(6,2),(6,5),(6,7),(6,8),\\(7,1),(7,2),(7,5),(7,7),(7,8),(8,1),(8,2),(8,5),(8,7),(8,8)\}[/tex]

The sum of the outcomes are:

[tex]S = \{2,3,6,8,9,3,4,7,9,10,4,5,8,10,11,5,6,9,11,12,6,7,10,12,\\13,7,8,11,13,14,8,9,12,14,15,9,10,13,15,16\}[/tex]

Rearrange

[tex]S = \{2,3,3,4,4,5,5,6,6,6,7,7,7,8,8,8,8,9,9,9,9,9,10,10,10,10,11,11,11,\\12,12,12,13,13,13,14,14,15,15,16\}[/tex]

[tex]n(S) = 40[/tex]

Next, we test the options

(1) The game is unfair

The question implies that the eight sided die is fair. This means that the game is fair.

Option (1) is false.

(2) Probability that multiple of 3 is [tex]\frac{3}{8}[/tex]

The multiples of 3 in the sample space are:

[tex]3's = \{3,3,6,6,6,9,9,9,9,9,12,12,12,15,15\}[/tex]

[tex]n(3's) =15[/tex]

The probability is then calculated as:

[tex]P(3's) = \frac{n(3's)}{n(S)}[/tex]

[tex]P(3's) = \frac{15}{40}[/tex]

Simplify

[tex]P(3's) = \frac{3}{8}[/tex]

Option (2) is true

(3) Probability of even sum is [tex]\frac{1}{2}[/tex]

The even outcomes are:

[tex]Even =\{2,4,4,6,6,6,8,8,8,8,10,10,10,10,12,12,12,14,14,16\}[/tex]

[tex]n(Even) = 20[/tex]

The probability is then calculated as:

[tex]P(Even)= \frac{n(Even)}{n(S)}[/tex]

[tex]P(Even)= \frac{20}{40}[/tex]

Simplify

[tex]P(Even)= \frac{1}{2}[/tex]

Option (3) is true

(4) Probability of a sum greater than or equal to 12 is [tex]\frac{11}{40}[/tex]

The outcomes of a sum greater than or equal to 12 are:

[tex]Sum \ge 12 = \{12,12,12,13,13,13,14,14,15,15,16\}[/tex]

[tex]n(Sum \ge 12) = 11[/tex]

The probability is then calculated as:

[tex]P(Sum \ge 12) =\frac{n(P(Sum \ge 12)}{n(S)}[/tex]

[tex]P(Sum \ge 12) =\frac{11}{40}[/tex]

Option (4) is true

(5) Probability of a sum less than 10 is [tex]\frac{21}{40}[/tex]

The outcomes of a sum less than 10 are:

[tex]Sum < 10 = \{2,3,3,4,4,5,5,6,6,6,7,7,7,8,8,8,8,9,9,9,9,9\}[/tex]

[tex]n(Sum < 10) = 22[/tex]

The probability is then calculated as:

[tex]P(Sum < 10) =\frac{n(P(Sum < 10)}{n(S)}[/tex]

[tex]P(Sum < 10) =\frac{22}{40}[/tex]

Option (5) is false

Hence, the true statements are: options 2, 3 and 4.

Read more about probabilities at:

https://brainly.com/question/20954641

Answer:

2.) The probability of getting a sum that is a multiple of 3 is 3/8

3.) The probability of getting a sum that is even is 1/2.

4.) The probability of getting a sum that is greater than or equal to 12 is 11/40

Explanation:

X = 10
3.2 + 5y = 20
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