You are playing a game that involves rolling two fair dice (number cubes) numbered 1 through 6.
If the score for one roll of the game is the sum of the two numbers on the dice, what are the smallest and largest possible outcomes? What is the expected value for the outcome? What is the value and the probability the most likely outcome? Show your work!
Thanks so much for your help.
The smallest outcome is: 1 + 1 = 2
The largest outcome is: 6 + 6 = 12
The expected value: [1+2+3+4+5+6] x 2 / 6 = 7
The value of the most likely outcome is 7 as above:
The probability of getting a 7 is: [1+6, 2+5, 3+4, 4+3, 5+2, 6+1] =6 outcomes / 6^2 = 6/36 =1/6
+983 1. All we need to do is find the expected value for one die, then do what the problem asks us to do.
The expected value for the first die was:
\(\frac{1+2+3+4+5+6}{6}=3.5\),
\(3.5+3.5=\boxed7\),
That is the expected value for the outcome.
2.
The probabilty of getting a sum of 2 is: \(\frac{1}{36}\). The numerator is one, because there is one way of getting a sum of 2, 1 + 1.
We do the same for the rest of the sums,
The probabilty of getting a sum of 3 is: \(\frac{2}{36}\), (1,2) ; (2,1)
The probabilty of getting a sum of 4 is: \(\frac{3}{36}\), (1,3) ; (3,1) ; (2,2)
The probabilty of getting a sum of 5 is: \(\frac{4}{36}\), (1,4) ; (4,1) ; (2,3) ; (3,2)
The probabilty of getting a sum of 6 is: \(\frac{5}{36}\), (1,5) ; (5,1) ; (2,4) ; (4,2) ; (3,3)
The probabilty of getting a sum of 7 is: \(\frac{6}{36}\), (1,6) ; (6,1) ; (2,5) ; (5,2) ; (3,4) ; (4,3)
The probabilty of getting a sum of 8 is: \(\frac{5}{36}\), (2,6) ; (6,2) ; (3,5) ; (5,3) ; (4,4)
The probabilty of getting a sum of 9 is: \(\frac{4}{36}\), (3,6) ; (6,3) ; (4,5) ; (5,4)
The probabilty of getting a sum of 10 is: \(\frac{3}{36}\), (4,6) ; (6,4) ; (5,5)
The probabilty of getting a sum of 11 is: \(\frac{2}{36}\), (5,6) ; (6,5)
The probabilty of getting a sum of 12 is: \(\frac{1}{36}\), (6,6)
The most likely outcome is getting a 7, with a 1/6 probability.
3.
Greatest is 6 + 6 = 12
Smallest is 1 + 1 = 2
