4.6 #94 Two fair dice are rolled. The sample space S contains the combinations o

Question

4.6 #94

Two fair dice are rolled. The sample space S contains the combinations of two numbers. For each mebmer (r,s) of S, the random variable X(r,s) = r + s.

a. Write a table showing the values for X and the probability of those values; instead of 36 columns each with probability 1/36, do a column for each distinct value of X and show the probability of that value.

b. Find the expected value of the sum of the numbers that come up when two fair dice are rolled.

c. Find the expected value of the sum of the numbers that come up when two fair dice are rolled. This time let the sample space S consist of the ordered pairs (r,s) that can appear on the two dice. Use two different random variables over this sample space, where X1 = the value of the first component of the order pair and X2 =the value of the second component of the ordered pair. Make use of the linearity property from Equation (4).

Explanation / Answer

Here we have

Sum related ordered pairs

2 (1,1)

3 (2,1)(1,2)

4 (3,1)(1,3)(2,2)

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

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

7 (3,4),(4,3),(5,2),(2,5),(6,1),(1,6)

8 (5,3),(3,5),(6,2),(2,6),(4,4),

9 (5,4),(4,5),(6,3),(3,6)

10 (5,5),(6,4),(4,6)

11 (5,6),(6,5)

12 (6,6)

So related table will be given as :

X(r,s)=r+s | P(r,s)

----------------|-----------------------

2 | 1/36

3 | 2/36

4 | 3/36

5 | 4/36

6 | 5/36

7 | 6/36

8 | 5/36

9 | 4/36

10 | 3/36

11 | 2/36

12 | 1/36

This is the answer of part (a)

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Now as the E(X) is the sum of related X and P(x), so required

E(X)= 2(1/36) + 3(2/36) + 4(3/36) + 5(4/36) + 6(5/36) + 7(6/36)+ 8(5/36) + 9(4/36)+ 10(3/36) +11(2/36)+12(1/36)

=2/36+ 6/36 + 12/36 + 20/36 + 30/36 + 42/36 + 40/36 + 36/36 + 30/36 + 22/36 + 12/36

=252/36 = 7

that is the answer of part (b) and part (c)

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