Two students have devised a dice game named “Sums” for their statistics class. The game consists of choosing to play odds or evens.
Probabilities for “Sums”
Roll
2
3
4
5
6
7
8
9
10
11
12
P(roll)
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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?
Answers
Probability of getting points in odd & even are equal , hence Jessica can choose any one to play odd or even
Step-by-step explanation:
two dice can be thrown in 6 * 6 = 36 Ways
Sum
2 = ( 1 , 1) = 1/36
3 = (1 ,2 ) , (2 ,1 ) = 2/36
4 = (1 , 3) , (2 , 2) , ( 3, 1) = 3/36
5 = (1 ,4) , ( 2, 3) , ( 3, 2) , ( 4 ,1 ) = 4/36
6 = ( 1. 5) . (2 , 3) , ( 3, 3) , (4 . 2) , (5 , 1) = 5/36
7 = ( 1. 6) . (2 , 5) ( 3, 4) , ( 4 , 3) , (5 2) , ( 6 , 1) = 6/36
8= ( 2 , 6) , (3 , 5) , ( 4, 4) , ( 5 ,3) , (6 , 2) = 5/36
9 = (3 , 6) , ( 4 , 5) , ( 5 , 4 ) , (6 , 3) = 4/36
10 = (4 , 6 ) , (5 , 5) , ( 6. 4) = 3/36
11 = (5 ,6) , (6 , 5) = 2/36
12 = (6 , 6) = 1/36
Points odd number will get
= 3(2/36) + 5(4/36) + 7(6/36) + 9(4/36) + 11(2/36)
=( 6 + 20 + 42 + 36 + 22)/36
= 126/36
= 7/2
Points Even number will get
= 2(1/36) + 4(3/36) + 6(5/36) + 8(5/36) + 10(3/36) + 12(1/36)
=( 2 + 12 + 30 + 40 + 30 + 12)/36
= 126/36
= 7/2
Probability of getting points in odd & even are equal
hence Jessica can choose any one to play odd or even
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