Question

1. kelly rolls two fair six - sided dice and records the sum of the two numbers.

a. complete the table below showing the sample space of all possible sums she could get when rolling two dice.
b. what is the probability of getting a sum of 7?
c. what is the probability of getting a sum of 9 or higher?
d. what is the probability of getting a sum that is even?
e. define two events that have the same probability.
f. are all outcomes in the sample space equally likely? explain.
g. are all sums of the two dice equally likely? explain.

Explanation:

Step1: Calculate total number of outcomes

When rolling two six - sided dice, the total number of outcomes is \(n(S)=6\times6 = 36\) (by the fundamental counting principle).

Step2: Find number of outcomes for sum of 7

From the table, the pairs \((1,6)\), \((2,5)\), \((3,4)\), \((4,3)\), \((5,2)\), \((6,1)\) give a sum of 7. So \(n(7)=6\).
The probability \(P(7)=\frac{n(7)}{n(S)}=\frac{6}{36}=\frac{1}{6}\).

Step3: Find number of outcomes for sum of 9 or higher

For sum of 9: \((3,6)\), \((4,5)\), \((5,4)\), \((6,3)\) (\(n(9) = 4\)); for sum of 10: \((4,6)\), \((5,5)\), \((6,4)\) (\(n(10)=3\)); for sum of 11: \((5,6)\), \((6,5)\) (\(n(11) = 2\)); for sum of 12: \((6,6)\) (\(n(12)=1\)).
\(n(\geq9)=4 + 3+2 + 1=10\).
The probability \(P(\geq9)=\frac{n(\geq9)}{n(S)}=\frac{10}{36}=\frac{5}{18}\).

Step4: Find number of outcomes for even sum

Sum of 2: \((1,1)\) (\(n(2)=1\)); sum of 4: \((1,3)\), \((2,2)\), \((3,1)\) (\(n(4)=3\)); sum of 6: \((1,5)\), \((2,4)\), \((3,3)\), \((4,2)\), \((5,1)\) (\(n(6)=5\)); sum of 8: \((2,6)\), \((3,5)\), \((4,4)\), \((5,3)\), \((6,2)\) (\(n(8)=5\)); sum of 10: \(n(10) = 3\); sum of 12: \(n(12)=1\).
\(n(\text{even})=1+3 + 5+5+3+1=18\).
The probability \(P(\text{even})=\frac{n(\text{even})}{n(S)}=\frac{18}{36}=\frac{1}{2}\).

Step5: Define two events with same probability

Event \(A\): sum of 2 (\(n(2) = 1\)); Event \(B\): sum of 12 (\(n(12)=1\)). \(P(A)=P(B)=\frac{1}{36}\).

Step6: Check if all outcomes in sample space are equally likely

Each pair \((i,j)\) where \(i,j\in\{1,2,\cdots,6\}\) has a probability of \(\frac{1}{36}\). So, yes, all outcomes in the sample space are equally likely.

Step7: Check if all sums are equally likely

Since \(n(2)=1\), \(n(3)=2\), \(n(4)=3\), \(n(5)=4\), \(n(6)=5\), \(n(7)=6\), \(n(8)=5\), \(n(9)=4\), \(n(10)=3\), \(n(11)=2\), \(n(12)=1\), all sums are not equally likely.

Answer:

b. \(\frac{1}{6}\)
c. \(\frac{5}{18}\)
d. \(\frac{1}{2}\)
e. Event \(A\): sum of 2; Event \(B\): sum of 12 (answers may vary)
f. Yes, each outcome \((i,j)\) has probability \(\frac{1}{36}\)
g. No, different sums have different number of pairs.