Answer :
Let's analyze Jessica's situation in the dice game "Sums" to determine whether she should choose to play odds or evens.
We are given the probabilities associated with each of the possible sums from rolling two dice, as shown in the table:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{Roll} & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline P(\text{Roll}) & \frac{1}{36} & \frac{2}{36} & \frac{3}{36} & \frac{4}{36} & \frac{5}{36} & \frac{6}{36} & \frac{5}{36} & \frac{4}{36} & \frac{3}{36} & \frac{2}{36} & \frac{1}{36} \\ \hline \end{array} \][/tex]
We will calculate the expected value of points for both the even and odd sums.
### Even Sums and Their Probabilities
The even sums are [tex]$2, 4, 6, 8, 10, 12$[/tex]. From the table, we extract their probabilities:
[tex]\[ \begin{aligned} P(2) &= \frac{1}{36},\\ P(4) &= \frac{3}{36},\\ P(6) &= \frac{5}{36},\\ P(8) &= \frac{5}{36},\\ P(10) &= \frac{3}{36},\\ P(12) &= \frac{1}{36}. \end{aligned} \][/tex]
Next, we calculate the expected value of points for even sums, \(E(\text{evens})\):
[tex]\[ E(\text{evens}) = 2 \cdot \frac{1}{36} + 4 \cdot \frac{3}{36} + 6 \cdot \frac{5}{36} + 8 \cdot \frac{5}{36} + 10 \cdot \frac{3}{36} + 12 \cdot \frac{1}{36} \][/tex]
[tex]\[ E(\text{evens}) = \frac{2}{36} + \frac{12}{36} + \frac{30}{36} + \frac{40}{36} + \frac{30}{36} + \frac{12}{36} = \frac{126}{36} = 3.5 \][/tex]
### Odd Sums and Their Probabilities
The odd sums are [tex]$3, 5, 7, 9, 11$[/tex]. From the table, we extract their probabilities:
[tex]\[ \begin{aligned} P(3) &= \frac{2}{36},\\ P(5) &= \frac{4}{36},\\ P(7) &= \frac{6}{36},\\ P(9) &= \frac{4}{36},\\ P(11) &= \frac{2}{36}. \end{aligned} \][/tex]
Next, we calculate the expected value of points for odd sums, \(E(\text{odds})\):
[tex]\[ E(\text{odds}) = 3 \cdot \frac{2}{36} + 5 \cdot \frac{4}{36} + 7 \cdot \frac{6}{36} + 9 \cdot \frac{4}{36} + 11 \cdot \frac{2}{36} \][/tex]
[tex]\[ E(\text{odds}) = \frac{6}{36} + \frac{20}{36} + \frac{42}{36} + \frac{36}{36} + \frac{22}{36} = \frac{126}{36} = 3.5 \][/tex]
### Conclusion
The expected value of points for both even and odd sums is the same, \(3.5\).
So, the accurate statement in guiding Jessica is:
Both evens and odds have the same expected value of points, [tex]\(3.5\)[/tex]. Therefore, it makes no statistical difference whether Jessica chooses to play evens or odds, as her expected score will be the same in either case.
We are given the probabilities associated with each of the possible sums from rolling two dice, as shown in the table:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{Roll} & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline P(\text{Roll}) & \frac{1}{36} & \frac{2}{36} & \frac{3}{36} & \frac{4}{36} & \frac{5}{36} & \frac{6}{36} & \frac{5}{36} & \frac{4}{36} & \frac{3}{36} & \frac{2}{36} & \frac{1}{36} \\ \hline \end{array} \][/tex]
We will calculate the expected value of points for both the even and odd sums.
### Even Sums and Their Probabilities
The even sums are [tex]$2, 4, 6, 8, 10, 12$[/tex]. From the table, we extract their probabilities:
[tex]\[ \begin{aligned} P(2) &= \frac{1}{36},\\ P(4) &= \frac{3}{36},\\ P(6) &= \frac{5}{36},\\ P(8) &= \frac{5}{36},\\ P(10) &= \frac{3}{36},\\ P(12) &= \frac{1}{36}. \end{aligned} \][/tex]
Next, we calculate the expected value of points for even sums, \(E(\text{evens})\):
[tex]\[ E(\text{evens}) = 2 \cdot \frac{1}{36} + 4 \cdot \frac{3}{36} + 6 \cdot \frac{5}{36} + 8 \cdot \frac{5}{36} + 10 \cdot \frac{3}{36} + 12 \cdot \frac{1}{36} \][/tex]
[tex]\[ E(\text{evens}) = \frac{2}{36} + \frac{12}{36} + \frac{30}{36} + \frac{40}{36} + \frac{30}{36} + \frac{12}{36} = \frac{126}{36} = 3.5 \][/tex]
### Odd Sums and Their Probabilities
The odd sums are [tex]$3, 5, 7, 9, 11$[/tex]. From the table, we extract their probabilities:
[tex]\[ \begin{aligned} P(3) &= \frac{2}{36},\\ P(5) &= \frac{4}{36},\\ P(7) &= \frac{6}{36},\\ P(9) &= \frac{4}{36},\\ P(11) &= \frac{2}{36}. \end{aligned} \][/tex]
Next, we calculate the expected value of points for odd sums, \(E(\text{odds})\):
[tex]\[ E(\text{odds}) = 3 \cdot \frac{2}{36} + 5 \cdot \frac{4}{36} + 7 \cdot \frac{6}{36} + 9 \cdot \frac{4}{36} + 11 \cdot \frac{2}{36} \][/tex]
[tex]\[ E(\text{odds}) = \frac{6}{36} + \frac{20}{36} + \frac{42}{36} + \frac{36}{36} + \frac{22}{36} = \frac{126}{36} = 3.5 \][/tex]
### Conclusion
The expected value of points for both even and odd sums is the same, \(3.5\).
So, the accurate statement in guiding Jessica is:
Both evens and odds have the same expected value of points, [tex]\(3.5\)[/tex]. Therefore, it makes no statistical difference whether Jessica chooses to play evens or odds, as her expected score will be the same in either case.
