+118677 They are not hard to count.
|a-b|=c
a and b are interchangeable so there are really twice as many oucomes in the sample space
So that is 30 possible outcomes
Count those with a 3 and double it as well. 7*2=14 favourable outcomes
Prob is 14/30 = 7/15
| a | b | c | |
| 6 | 5 | 1 | |
| 5 | 4 | 1 | |
| 3 | 4 | 3 | 1 |
| 3 | 3 | 2 | 1 |
| 2 | 1 | 1 | |
| 6 | 4 | 2 | |
| 3 | 5 | 3 | 2 |
| 4 | 2 | 2 | |
| 3 | 3 | 1 | 2 |
| 3 | 6 | 3 | 3 |
| 3 | 5 | 2 | 3 |
| 3 | 4 | 1 | 3 |
| 6 | 2 | 4 | |
| 5 | 1 | 4 | |
| 6 | 1 | 5 |
+26393 A fair six-sided number cube with the digits 1-6 on its faces is rolled three times.
The positive difference between the values on the first two rolls is equal to the value of the third roll.
What is the probability that at least one 3 was rolled?
Express your answer as a common fraction.
\(\begin{array}{|c|c|c|c|} \hline \text{first roll} & \text{second roll} & \text{third roll} \\ \mathbf{ 1} & (\text{difference}) &\text{1 2 $\color{red}3$ 4 5 6} \\ \hline & 1~(0) & \text{_ _ _ _ _ _} \\ & 2~(1) & \text{$\color{blue}x$_ _ _ _ _} \\ & {\color{red}3}~(2) & \text{_ $\color{red}x$ _ _ _ _} \\ & 4~(3) & \text{_ _ $\color{red}x$ _ _ _} \\ & 5~(4) & \text{_ _ _ $\color{blue}x$ _ _} \\ & 6~(5) & \text{_ _ _ _ $\color{blue}x$ _} \\ \hline \end{array} \begin{array}{|c|c|c|c|} \hline \text{first roll} & \text{second roll} & \text{third roll} \\ \mathbf{2} & (\text{difference}) &\text{1 2 $\color{red}3$ 4 5 6} \\ \hline & 1~(1) & \text{$\color{blue}x$_ _ _ _ _} \\ & 2~(0) & \text{_ _ _ _ _ _} \\ & {\color{red}3}~(1) & \text{$\color{red}x$_ _ _ _ _} \\ & 4~(2) & \text{_ $\color{blue}x$ _ _ _ _} \\ & 5~(3) & \text{_ _ $\color{red}x$ _ _ _} \\ & 6~(4) & \text{_ _ _ $\color{blue}x$ _ _} \\ \hline \end{array}\)
\(\begin{array}{|c|c|c|c|} \hline \text{first roll} & \text{second roll} & \text{third roll} \\ \mathbf{ {\color{red}3}} & (\text{difference}) &\text{1 2 $\color{red}3$ 4 5 6} \\ \hline & 1~(2) & \text{_ $\color{red}x$ _ _ _ _} \\ & 2~(1) & \text{$\color{red}x$_ _ _ _ _} \\ & {\color{red}3}~(0) & \text{_ _ _ _ _ _} \\ & 4~(1) & \text{$\color{red}x$_ _ _ _ _} \\ & 5~(2) & \text{_ $\color{red}x$ _ _ _ _} \\ & 6~(3) & \text{_ _ $\color{red}x$ _ _ _} \\ \hline \end{array} \begin{array}{|c|c|c|c|} \hline \text{first roll} & \text{second roll} & \text{third roll} \\ \mathbf{ 4} & (\text{difference}) &\text{1 2 $\color{red}3$ 4 5 6} \\ \hline & 1~(3) & \text{_ _ $\color{red}x$ _ _ _} \\ & 2~(2) & \text{_ $\color{blue}x$ _ _ _ _} \\ & {\color{red}3}~(1) & \text{$\color{red}x$_ _ _ _ _} \\ & 4~(0) & \text{_ _ _ _ _ _} \\ & 5~(1) & \text{$\color{blue}x$_ _ _ _ _} \\ & 6~(2) & \text{_ $\color{blue}x$ _ _ _ _} \\ \hline \end{array}\)
\(\begin{array}{|c|c|c|c|} \hline \text{first roll} & \text{second roll} & \text{third roll} \\ \mathbf{ 5} & (\text{difference}) &\text{1 2 $\color{red}3$ 4 5 6} \\ \hline & 1~(4) & \text{_ _ _ $\color{blue}x$ _ _} \\ & 2~(3) & \text{_ _ $\color{red}x$ _ _ _} \\ & {\color{red}3}~(2) & \text{_ $\color{red}x$ _ _ _ _} \\ & 4~(1) & \text{$\color{blue}x$_ _ _ _ _} \\ & 5~(0) & \text{_ _ _ _ _ _} \\ & 6~(1) & \text{$\color{blue}x$_ _ _ _ _} \\ \hline \end{array} \begin{array}{|c|c|c|c|} \hline \text{first roll} & \text{second roll} & \text{third roll} \\ \mathbf{ 6} & (\text{difference}) &\text{1 2 $\color{red}3$ 4 5 6} \\ \hline & 1~(5) & \text{_ _ _ _ $\color{blue}x$ _} \\ & 2~(4) & \text{_ _ _ $\color{blue}x$ _ _} \\ & {\color{red}3}~(3) & \text{_ _ $\color{red}x$ _ _ _} \\ & 4~(2) & \text{_ $\color{blue}x$ _ _ _ _} \\ & 5~(1) & \text{$\color{blue}x$_ _ _ _ _} \\ & 6~(0) & \text{_ _ _ _ _ _} \\ \hline \end{array}\)
The probability that at least one 3 was rolled is \(\mathbf{ \dfrac{14\ {\color{red}x} }{30\ {\color{blue}x} } = \dfrac{7}{15} }\)
Answer: 7/15
