Question

b. one way to analyze this situation is to make a table of all the possible outcomes like the one below. copy and complete this table of sums on your paper. are each of the outcomes in this table equally likely?

c. what is ( p(\text{even}) )? ( p(10) )? ( p(15) )?

d. which sum is the most likely result? what is the probability of rolling that s

Explanation:

Step1: Complete the table

For each cell in the table, the value is the sum of the row - label (Dice #2) and the column - label (Dice #1).

• When Dice #2 = 1 and Dice #1 = 1: \(1 + 1=2\)
• When Dice #2 = 1 and Dice #1 = 2: \(1 + 2 = 3\)
• When Dice #2 = 1 and Dice #1 = 3: \(1+3 = 4\)
• When Dice #2 = 1 and Dice #1 = 5: \(1 + 5=6\)
• When Dice #2 = 1 and Dice #1 = 6: \(1+6 = 7\)
• When Dice #2 = 2 and Dice #1 = 1: \(2 + 1=3\)
• When Dice #2 = 2 and Dice #1 = 3: \(2+3 = 5\)
• When Dice #2 = 2 and Dice #1 = 4: \(2 + 4=6\)
• When Dice #2 = 2 and Dice #1 = 5: \(2+5 = 7\)
• When Dice #2 = 2 and Dice #1 = 6: \(2+6 = 8\)
• When Dice #2 = 3 and Dice #1 = 1: \(3 + 1=4\)
• When Dice #2 = 3 and Dice #1 = 2: \(3+2 = 5\)
• When Dice #2 = 3 and Dice #1 = 4: \(3 + 4=7\)
• When Dice #2 = 3 and Dice #1 = 5: \(3+5 = 8\)
• When Dice #2 = 3 and Dice #1 = 6: \(3+6 = 9\)
• When Dice #2 = 4 and Dice #1 = 1: \(4 + 1=5\)
• When Dice #2 = 4 and Dice #1 = 2: \(4+2 = 6\)
• When Dice #2 = 4 and Dice #1 = 5: \(4 + 5=9\)
• When Dice #2 = 4 and Dice #1 = 6: \(4+6 = 10\)
• When Dice #2 = 5 and Dice #1 = 1: \(5 + 1=6\)
• When Dice #2 = 5 and Dice #1 = 2: \(5+2 = 7\)
• When Dice #2 = 5 and Dice #1 = 3: \(5 + 3=8\)
• When Dice #2 = 5 and Dice #1 = 4: \(5+4 = 9\)
• When Dice #2 = 5 and Dice #1 = 6: \(5+6 = 11\)
• When Dice #2 = 6 and Dice #1 = 1: \(6 + 1=7\)
• When Dice #2 = 6 and Dice #1 = 2: \(6+2 = 8\)
• When Dice #2 = 6 and Dice #1 = 3: \(6 + 3=9\)
• When Dice #2 = 6 and Dice #1 = 4: \(6+4 = 10\)
• When Dice #2 = 6 and Dice #1 = 5: \(6+5 = 11\)
• When Dice #2 = 6 and Dice #1 = 6: \(6+6 = 12\)

The completed table:

Dice #2\Dice #1	1	2	3	4	5	6
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

Since each combination of Dice #1 and Dice #2 (e.g., (1,1),(1,2),\(\cdots\),(6,6)) has a probability of \(\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}\), each outcome in the table is equally likely.

Step2: Calculate \(P(\text{even})\)

The total number of outcomes \(n(S)=36\).
The number of even - sum outcomes:
The even sums are 2,4,6,8,10,12.

• For sum = 2: 1 outcome \((1,1)\)
• For sum = 4: 3 outcomes \((1,3),(2,2),(3,1)\)
• For sum = 6: 5 outcomes \((1,5),(2,4),(3,3),(4,2),(5,1)\)
• For sum = 8: 5 outcomes \((2,6),(3,5),(4,4),(5,3),(6,2)\)
• For sum = 10: 3 outcomes \((4,6),(5,5),(6,4)\)
• For sum = 12: 1 outcome \((6,6)\)

The number of even - sum outcomes \(n(\text{even})=1 + 3+5+5+3 + 1=18\)
\(P(\text{even})=\frac{n(\text{even})}{n(S)}=\frac{18}{36}=\frac{1}{2}\)

Step3: Calculate \(P(10)\)

The number of outcomes with sum = 10: 3 outcomes \((4,6),(5,5),(6,4)\)
\(P(10)=\frac{3}{36}=\frac{1}{12}\)

Step4: Calculate \(P(15)\)

Since the maximum sum of two six - sided dice is \(6 + 6=12\), \(n(15)=0\)
\(P(15)=\frac{0}{36}=0\)

Step5: Find the most - likely sum

The sum of 7 has the most number of outcomes.
The number of outcomes with sum = 7: 6 outcomes \((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\)
\(P(7)=\frac{6}{36}=\frac{1}{6}\)

Answer:

• The completed table:

Dice #2\Dice #1	1	2	3	4	5	6
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

Each outcome in the table is equally likely.

• \(P(\text{even})=\frac{1}{2}\), \(P(10)=\frac{1}{12}\), \(P(15)=0\)
• The most - likely sum is 7 and \(P(7)=\frac{1}{6}\)