In a game three standard dice are rolled and the number of o

In a game, three standard dice are rolled and the number of odd values that appear is used to advance your gamepiece (for example, the roll 2-3-1 would advance your gamepiece two spaces). Produce a probability distribution for this experiment.

Compute the mean and standard deviation of the probability distribution in the first DO IT NOW problem

Solution

There are 6×6×6 = 216 ways the three dice can fall.

The number of ways to get 3 evens is 3×3×3 = 27, in which would advance your gamepiece 0 spaces.

The number of ways to get 3 odds is also 3×3×3 = 27, in which would advance your gamepiece 3 spaces.

That accounts for 27+27 or 54 of the 216 rolls. That leaves 216-54=162 possible rolls.

The remaining 162 rolls will consist of either 2 odds and 1 even,or 2 evens and 1 odd.

There will be the same number of each, so there will be 162/2 = 81 ways to roll 2 odds and 1 even, in which would advance the gamepiece 2 spaces and there will be 162/2 = 81 ways to roll 2 evens and 1 odd, in which would advance the gamepiece 1 spaces So,

the probability of moving the gamepiece 0 spaces is 27/216 = 1/8 = 0.125

the probability of moving the gamepiece 1 spaces is 81/216 = 3/8 = 0.375

the probability of moving the gamepiece 2 spaces is 81/216 = 3/8 = 0.375

the probability of moving the gamepiece 3 spaces is 81/216 = 1/8 = 0.125

X           P(X)       Expect-

[No. of      [Prob-      ation

spaces        abil-        =

moved]        ity]       X·P(X)

-------------------------------

0          0.125          0

1          0.375        0.375   

2          0.375        0.75

3          0.375        1.125

-------------------------------

             1.000        2.25

The expectation of 2.25 means that if we roll the three dice many times, will average moving the gamepiece 2.25 spaces per roll. [we will never move it 2.25 spaces, because we will either move it 0,1,2, or 3 spaces. However we will average moving it 2.25 spaces per roll].