Consider rolling a fair die twice, and let S denote all combinations of possible

Question

Consider rolling a fair die twice, and let S denote all combinations of possible outcomes from the two roll S = {(1, 1), (1,2), (1,6), (2, 1), (2,2), (6,6)} Suppose that any outcome in S is equally likely, P[(i.j)] - Pl(',j')] for all (i, j), (i, j') S. Let X denote the sum of the two numbers from rolling the die twice. (a) Find P(i, j)| for each (i, j) S. (b) Find PXr for r E {2,3, ..., 12). (c) What is PX for r2,3, ..., 12)? (d) Which value of X is the most likely? in other words, find the value r* such that P[X-r] 2 PXr for all . Give an intuitive explanation for why that value is the most likely. (e) Let F(t) = P(X-t) denote the cumulative distribution function of X Find F(t) for all t (f Graph F(t). (g) Let E(X) 122xPX 2]. Find E(X) (h) How does E(X) compare to the a" you found in part (d)? Is E(X) big- ger then, smaller than, or equal to from part (d)? Give an intuitive explanation for the relationship between E(X) and x* in this problem

Explanation / Answer

Given

S = { (1,1),(1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

n = 36

a) P[(i,j)] = 1/36 for each (i,j) belongs to S

b) Let X denote sum of the numbers on two dice

Sum (x): 2 3 4 5 6 7 8 9 10 11 12

Prob: 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

c) P(X=x) = 0 for x not belongs to {2,3,...12}

d) We find the value of x* such that P(X=x*) >= P(X=x)

From part (b) , x* = 7

e) The CDF of X is

Sum (x): 2 3 4 5 6 7 8 9 10 11 12

Prob: 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
F(t) : 1/36 3/36 6/36 10/36 15/36 21/36 26/36 30/36 33/36 35/36 36/36

g)

E(X) = 2(1/36) + 3(2/36) + 4(3/26) + 5 (4/36) + 6(5/36) + 7(6/36) + 8(5/36) + 9(3/36) + 10(3/36) + 22(2/36) + 12(1/36) = 7

h) E(X) = 7 = x*

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