Question

1. roll and win

you begin the game roll and win by picking a number. then you roll two regular dice, each numbered 1 through 6, and add the numbers that come up together. if the sum is the number you chose, you win a point. for example, if you choose \11\, and a 5 and a 6 are rolled, you win!
a. what is the sample space, which can be thought of as the set of all the possible outcomes, when two dice are rolled and their numbers added?
b. one way to analyze this situation is to make a table of all the possible outcomes like the one below. complete this table of sums on your paper. are each of the outcomes in this table equally likely? this is not a probability area model
c. what is p(even)? p(15)? p(10)?
d. which sum is the most likely result? what is the probability of rolling that sum?

Explanation:

Step1: Find the sample - space for part a

The minimum sum of two dice is \(1 + 1=2\) and the maximum sum is \(6+6 = 12\). So the sample space \(S=\{2,3,4,5,6,7,8,9,10,11,12\}\).

Step2: Complete the table for part b

Dice #1\Dice #2	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

The outcomes are not equally likely. For example, there is only one way to get a sum of 2 (\((1,1)\)) but six ways to get a sum of 7 (\((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\)).

Step3: Calculate probabilities for part c

The total number of possible outcomes when rolling two dice is \(n(S)=6\times6 = 36\).

• For \(P(\text{even})\): The even - sum outcomes:

Sum of 2: 1 way \((1,1)\); Sum of 4: 3 ways \((1,3),(2,2),(3,1)\); Sum of 6: 5 ways \((1,5),(2,4),(3,3),(4,2),(5,1)\); Sum of 8: 5 ways \((2,6),(3,5),(4,4),(5,3),(6,2)\); Sum of 10: 3 ways \((4,6),(5,5),(6,4)\); Sum of 12: 1 way \((6,6)\). The number of even - sum outcomes \(n(\text{even})=1 + 3+5 + 5+3 + 1=18\). So \(P(\text{even})=\frac{18}{36}=\frac{1}{2}\).

• For \(P(15)\): Since the maximum sum of two dice is 12, \(n(15) = 0\), so \(P(15)=0\).
• For \(P(10)\): The number of ways to get a sum of 10 is 3 (\((4,6),(5,5),(6,4)\)), so \(P(10)=\frac{3}{36}=\frac{1}{12}\).

Step4: Find the most likely sum and its probability for part d

The sum of 7 has the most number of combinations (\(n(7)=6\)). So the most likely sum is 7. The probability \(P(7)=\frac{6}{36}=\frac{1}{6}\).

Answer:

a. The sample space is \(S = \{2,3,4,5,6,7,8,9,10,11,12\}\).
b. The completed table is shown above. The outcomes are not equally likely.
c. \(P(\text{even})=\frac{1}{2}\), \(P(15)=0\), \(P(10)=\frac{1}{12}\).
d. The most likely sum is 7 and \(P(7)=\frac{1}{6}\).