Question

1. christy is going to roll a pair of six - sided dice.

a. create a list of the sample space for this situation.
b. if we add the numbers on each die, what is the most likely outcome?
c. use your sample space to find the probability of rolling doubles (the same number on both die)

Explanation:

Part a

Step1: Define sample space for two dice

When rolling two six - sided dice, each die can show a number from 1 to 6. The sample space \(S\) consists of all ordered pairs \((x,y)\) where \(x\) is the result of the first die and \(y\) is the result of the second die.
So the sample space is \(\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}\)

Step1: Calculate the sum for each pair

For each pair \((x,y)\) in the sample space, calculate \(x + y\):

• For \((1,1)\), sum \(= 2\); \((1,2)\) sum \(= 3\); \((1,3)\) sum \(= 4\); \((1,4)\) sum \(= 5\); \((1,5)\) sum \(= 6\); \((1,6)\) sum \(= 7\)
• For \((2,1)\) sum \(= 3\); \((2,2)\) sum \(= 4\); \((2,3)\) sum \(= 5\); \((2,4)\) sum \(= 6\); \((2,5)\) sum \(= 7\); \((2,6)\) sum \(= 8\)
• For \((3,1)\) sum \(= 4\); \((3,2)\) sum \(= 5\); \((3,3)\) sum \(= 6\); \((3,4)\) sum \(= 7\); \((3,5)\) sum \(= 8\); \((3,6)\) sum \(= 9\)
• For \((4,1)\) sum \(= 5\); \((4,2)\) sum \(= 6\); \((4,3)\) sum \(= 7\); \((4,4)\) sum \(= 8\); \((4,5)\) sum \(= 9\); \((4,6)\) sum \(= 10\)
• For \((5,1)\) sum \(= 6\); \((5,2)\) sum \(= 7\); \((5,3)\) sum \(= 8\); \((5,4)\) sum \(= 9\); \((5,5)\) sum \(= 10\); \((5,6)\) sum \(= 11\)
• For \((6,1)\) sum \(= 7\); \((6,2)\) sum \(= 8\); \((6,3)\) sum \(= 9\); \((6,4)\) sum \(= 10\); \((6,5)\) sum \(= 11\); \((6,6)\) sum \(= 12\)

Step2: Count the frequency of each sum

• Sum of 2: 1 time
• Sum of 3: 2 times
• Sum of 4: 3 times
• Sum of 5: 4 times
• Sum of 6: 5 times
• Sum of 7: 6 times
• Sum of 8: 5 times
• Sum of 9: 4 times
• Sum of 10: 3 times
• Sum of 11: 2 times
• Sum of 12: 1 time

The sum with the highest frequency is 7.

Step1: Identify the number of doubles

The doubles in the sample space are \((1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\). So there are 6 favorable outcomes.

Step2: Find the total number of outcomes

From the sample space in part (a), the total number of possible outcomes when rolling two dice is \(6\times6 = 36\).

Step3: Calculate the probability

The probability \(P\) of an event is given by the formula \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\). So \(P(\text{rolling doubles})=\frac{6}{36}=\frac{1}{6}\)

Answer:

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

Part b