Consider two 3sided fair dice with sides 123 The two dice ar

Consider two 3-sided fair dice with sides 1,2,3. The two dice are rolled independently 36 times each, and the sample average of the outcomes for each die is recorded. What is the probability that the difference in the sample averages of the two dice is larger than 0.2? Since there might be round off errors, choose the closest answer to yours.

multiple choice: 0.149, 0.927, 0.298, None of the above, 0.451, 0.345, 0.643

Solution

We are given that there are two dice with three sides.

The two dice are rolled independently 36 times each and the sample average of the outcomes for each die is recorded.

First die coutcome = {(1,1) (1,2) (1,3),(2,1) (2,2) (2,3) (3,1) (3,2) (3,3)}

SImilarly for the second die outcome =  {(1,1) (1,2) (1,3),(2,1) (2,2) (2,3) (3,1) (3,2) (3,3)}

There are 9 possible outcomes.

We have to take a sample average of the outcomes for each die is so the possible outcomes are,

Possible averages are { 1 , 1.5 , 2 , 2.5 , 3 }

We take the possible differences of that averages.

{ -2 , -1.5 , -1 , -0.5 , 0 , 0.5 , 1 , 1.5 , 2}

And we have to calculate the probability that the difference in the sample averages of the two dice is larger than 0.2.

Let us consider random variable X is diffrence in the sample averages.

So X can taake the value which are larger than 0.2 as 0.5 or 1 or 1.5 or 2 .

P(X > 0.2) = P(X = 0.5 or 1 or 1.5 or 2)

= number of favourable ways / total number of ways

= 4 / 9 = 0.44

Which is nearly equal to 0.451.