Question

2 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 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

The pairs \((x,y)\) such that \(x + y=7\) are \((1,6)\), \((2,5)\), \((3,4)\), \((4,3)\), \((5,2)\), \((6,1)\). 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 \(=9\): \((3,6)\), \((4,5)\), \((5,4)\), \((6,3)\) (\(n(9) = 4\))
For sum \(=10\): \((4,6)\), \((5,5)\), \((6,4)\) (\(n(10)=3\))
For sum \(=11\): \((5,6)\), \((6,5)\) (\(n(11) = 2\))
For sum \(=12\): \((6,6)\) (\(n(12)=1\))
\(n(9\ or\ higher)=4 + 3+2 + 1=10\)
The probability \(P(9\ or\ higher)=\frac{n(9\ or\ higher)}{n(S)}=\frac{10}{36}=\frac{5}{18}\)

Step4: Find number of outcomes for even sum

Let the first die be \(x\) and the second die be \(y\). \(x + y\) is even. If \(x\) is odd (\(x = 1,3,5\)) and \(y\) is odd (\(y = 1,3,5\)): \(n_{1}=3\times3 = 9\)
If \(x\) is even (\(x = 2,4,6\)) and \(y\) is even (\(y = 2,4,6\)): \(n_{2}=3\times3=9\)
\(n(even)=9 + 9=18\)
The probability \(P(even)=\frac{n(even)}{n(S)}=\frac{18}{36}=\frac{1}{2}\)

Step5: Define two events with same probability

Event \(A\): sum of \(2\) (\((1,1)\), \(n(A)=1\))
Event \(B\): sum of \(12\) (\((6,6)\), \(n(B)=1\))
\(P(A)=P(B)=\frac{1}{36}\)

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

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

Step7: Check if all sums are equally likely

Since \(P(2)=\frac{1}{36}\), \(P(3)=\frac{2}{36}\), \(P(4)=\frac{3}{36}\), \(P(5)=\frac{4}{36}\), \(P(6)=\frac{5}{36}\), \(P(7)=\frac{6}{36}\), \(P(8)=\frac{5}{36}\), \(P(9)=\frac{4}{36}\), \(P(10)=\frac{3}{36}\), \(P(11)=\frac{2}{36}\), \(P(12)=\frac{1}{36}\)
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, probabilities of sums like \(2\), \(3,\cdots,12\) are different.