Question

Two fair dice are rolled. What is the probability of
rolling doubles or a number that is evenly divisible by 3?

Answer 1

@satellite73
@Directrix

Answer 2

add and divide by 36

Answer 3

holy %^$# that is all wrong!

Answer 4

it says "doubles" not 7s

Answer 5

ok we start again
3 ways to get a 3
5 ways to get a 6
4 ways to get a 9
1 way to get a 12
and 6 ways to get doubles, but we have to subtract off the doubles that we counded
12 is double sixes so subtract 1
one way to get a 6 is (3,3) so we have to subtract another

Answer 6

Two fair dice are rolled. What is the probability of
a) rolling a total of 8?
b) rolling a total greater than 5?
c) rolling a 2, 4 times in a row?
d) rolling doubles or a number that is evenly divisible by 3?

Answer 7

so we add \(5+3+4+1-1-1=11\) and your answer is \(\frac{11}{36}\)

Answer 8

Alright, thanks man! You're a big help. I appreciate your assistance.

Answer 9

Is it rolling the number on one die that is divisible by 3 or the sum of the numbers that is divisible by 3?

Answer 10

If we're looking at the sums of the numbers being divisible by three, then this is the table I would use.
\[\begin{array}{|l|c|c|c|c|c|c|}
\hline
+ & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
1 & \color{red}{2} & \color{blue}{3} & 4 & 5 & \color{blue}{6} & 7 \\ \hline
2 & \color{blue}{3} & \color{red}{4} & 5 & \color{blue}{6} & 7 & 8 \\ \hline
3 & 4 & 5 & \color{purple}{6} & 7 & 8 & \color{blue}{9} \\ \hline
4 & 5 & \color{blue}{6} & 7 & \color{red}{8} & \color{blue}{9} & 10 \\ \hline
5 & \color{blue}{6} & 7 & 8 & \color{blue}{9} & \color{red}{10} & 11 \\ \hline
6 & 7 & 8 & \color{blue}{9} & 10 & 11 & \color{purple}{12} \\
\hline
\end{array} \\
\quad \mathbb{\text{Blue}}= \text{Divisible by 3},\quad \mathbb{\text{Red}} = \text{Double},\quad \mathbb{\text{Purple}} = \text{Both}\]

However, if we're looking at individual rolls being divisible by three, I would use:
\[\begin{array}{|l|c|c|c|c|c|c|}
\hline
~ & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
1 & \color{Red}X & ~ & \color{Blue}X & ~ & ~ & \color{Blue}X \\ \hline
2 & ~ & \color{Red}X & \color{Blue}X & ~ & ~ & \color{Blue}X \\ \hline
3 & \color{Blue}X & \color{Blue}X & \color{Purple}X & \color{Blue}X & \color{Blue}X & \color{Blue}X \\ \hline
4 & ~ & ~ & \color{Blue}X & \color{Red}X & ~ & \color{Blue}X \\ \hline
5 & ~ & ~ & \color{Blue}X & ~ & \color{Red}X & \color{Blue}X \\ \hline
6 & \color{Blue}X & \color{Blue}X & \color{Blue}X & \color{Blue}X & \color{Blue}X & \color{purple}X \\
\hline
\end{array} \\
\quad \mathbb{\text{Blue}}= \text{Divisible by 3},\quad \mathbb{\text{Red}} = \text{Double},\quad \mathbb{\text{Purple}} = \text{Both}\]

Answer 11

My guess would have been for "individual rolls being divisible by 3," just because they seem to mention when we're looking at the total...

Answer 12

wow thats nice!!