Question

Two students have devised a dice game named “Sums” for their statistics class. The game consists of choosing to play odds or evens. Roll 2 3 4 5 6 7 8 9 10 11 12 P(roll) StartFraction 1 Over 36 EndFraction StartFraction 2 Over 36 EndFraction StartFraction 3 Over 36 EndFraction StartFraction 4 Over 36 EndFraction StartFraction 5 Over 36 EndFraction StartFraction 6 Over 36 EndFraction StartFraction 5 Over 36 EndFraction StartFraction 4 Over 36 EndFraction StartFraction 3 Over 36 EndFraction StartFraction 2 Over 36 EndFraction StartFraction 1 Over 36 EndFraction Each person takes turns rolling two dice. If the sum is odd, the person playing odds gets points equal to the sum of the roll. If the sum is even, the person playing evens gets points equal to the sum of the roll. Note that the points earned is independent of who is rolling the dice. If Jessica is challenged to a game of Sums, which statement below is accurate in every aspect in guiding her to the correct choice of choosing to play odds or evens? E(evens) will be more because there are more even numbers that result from rolling two dice. Therefore, Jessica should play evens. E(odds) will be more because the probability for each odd number being rolled is greater. Therefore, Jessica should play odds. E(evens) will be more because the value of the even numbers on the dice are more. Therefore, Jessica should play evens. E(evens) = E(odds) because the different probabilities and values end up balancing out, creating a fair game. Therefore, Jessica may choose whichever she likes.

Answer 1

Answer:

D /Even-odds

Step-by-step explanation:

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Answer 2

Answer:

(D)E(evens) = E(odds) because the different probabilities and values end up balancing out, creating a fair game. Therefore, Jessica may choose whichever she likes.

Step-by-step explanation:

The table of the probability of rolling the sums is presented below.

$\left\begin{array}{ccccccccccccc} Roll&2&3&4&5&6&7&8&9&10&11&12\\\\ Prob&\frac{1}{36}&\frac{2}{36}&\frac{3}{36}&\frac{4}{36}&\frac{5}{36}&\frac{6}{36}&\frac{5}{36}&\frac{4}{36}&\frac{3}{36}&\frac{2}{36}&\frac{1}{36} \end{array}\right$

P(an even sum)

$=\frac{1}{36}+\frac{3}{36}+\frac{5}{36}+\frac{5}{36}+\frac{3}{36}+\frac{1}{36} \\\\=\frac{18}{36}$

Therefore, P(an odd sum) $=\frac{18}{36}$

Therefore, E(evens) = E(odds) because the different probabilities and values end up balancing out, creating a fair game. Therefore, Jessica may choose whichever she likes.