A game is played with the following rules. Two ordinary number cubes, each with six sides labeled 1 to 6, are rolled. After each
roll a player is given a certain number of points depending on the outcome. If the sum is 2 or 12, the player receives +6 points If the sum is 3 or 11, the player receives +2 points. If the sum is any other number, the player receives –1 point. What is the expected value of the number of points a player will receive after a roll?
1) Expected value: is the weighted average of the values, being the probabilities the weight.
That is: ∑ of prbability of event i × value of event i.
In this case: (probability of getting 2 or 12) × (+6) + (probability of gettin 3 or 11) × (+2) + (probability of any other sum) × (-1).
2) Sample space:
Sum Points awarded
1+ 1 = 2 +6
1 + 2 = 3 +2
1 + 3 = 4 -1
1 + 4 = 5 -1
1 + 5 = 6 -1
1 + 6 = 7 -1
2 + 1 = 3 +2
2 + 2 = 4 -1
2 + 3 = 5 -1
2 + 4 = 6 -1
2 + 5 = 7 -1
2 + 6 = 8 -1
3 + 1 = 4 -1
3 + 2 = 5 -1
3 + 3 = 6 -1
3 + 4 = 7 -1
3 + 5 = 8 -1
3 + 6 = 9 -1
4 + 1 = 5 -1
4 + 2 = 6 -1
4 + 3 = 7 -1
4 + 4 = 8 -1
4 + 5 = 9 -1
4 + 6 = 10 -1
5 + 1 = 6 -1
5 + 2 = 7 -1
5 + 3 = 8 -1
5 + 4 = 9 -1
5 + 5 = 10 -1
5 + 6 = 11 +2
6 + 1 = 7 -1
6 + 2 = 8 -1
6 + 3 = 9 -1
6 + 4 = 10 -1
6 + 5 = 11 +2
6 + 6 = 12 +6
2) Probabilities
From that, there is: - 2/36 probabilities to earn + 6 points. - 4/36 probabilites to earn + 2 points - the rest, 30/36 probabilities to earn - 1 points
P=3 and p=-1 First you subtract the number without a variable to the other side, so subtract one from seven, and subtract five from eight. After that just divide to make p by its self and you get the answer!!
