Two dice - probability that sum of scores > 9 or individual scores differ by 1.

• November 20th 2011, 06:04 PM
Punch
Two dice - probability that sum of scores > 9 or individual scores differ by 1.
Two fair dice are thrown together, and the scores added. What is the probability that the total score is 9, or the individual scores differ by 1, or both?

$P(total score is 9)=\frac{4}{36}=\frac{1}{9}$

$P(individual scores differ by 1)=P{(1,2), (2,3), (2,1), (3,4), (3,2), (4,5),
(4,3), (5,6), (5,4), (6,5)}=\frac{10}{36}=\frac{5}{18}$

$P(both scoring 9 and score differ by 1)=P{(4,5),(5,4)}=\frac{2}{36}=\frac{1}{18}$

Should i treat the question as three seperate parts? The model answer has only 1 answer of 1/3 and that doesn't coincide with any of my answers...
• November 20th 2011, 09:31 PM
Soroban
Re: Two dice - probability that sum of scores > 9 or individual scores differ by 1.
Hello, Punch!

Quote:

Two fair dice are thrown together, and the scores added.
What is the probability that the total score is 9,
or the individual scores differ by 1, or both?

There are only 36 possible outcomes.
Why not list them and count the desired outcomes?

. . $\begin{array}{cccccc}(1,1) & {\color{red}(1,2)} & (1,3) & (1,4) & (1,5) & (1,6) \\ {\color{red}(2,1)} & (2,2) & {\color{red}(2,3)} & (2,4) & (2,5) & (2,6) \\ (3,1) & {\color{red}(3,2)} & (3,3) & {\color{red}(3,4)} & (3,5) & {\color{blue}(3,6)} \\ (4,1) & (4,2) & {\color{red}(4,3)} & (4,4) & {\color{blue}(4,5)} & (4,6) \\ (5,1) & (5,2) & (5,3) & {\color{blue}(5,4)} & (5,5) & {\color{red}(5,6)} \\ (6,1) & (6,2) & {\color{blue}(6,3)} & (6,4) & {\color{red}(6,5)} & (6,6) \end{array}$

There are $12$ desired outcomes.

The probability is:. $\frac{12}{36} \,=\,\frac{1}{3}$

• November 20th 2011, 10:35 PM
takatok
Re: Two dice - probability that sum of scores > 9 or individual scores differ by 1.
While enumerating every possibility and finding the answer by counting is possible, it can become unwieldy for large cases. For example what if we were using 100 sided dice instead of 6. You were working on the correct solution you just missed it by one step. I will use your probabilities since they are correct.

$P(9) = \frac{1}{9} = \frac{2}{18}$

$P(diff1) =\frac{5}{18}$

So to get the probability of both we just add them to get $\frac{7}{18}$

This is a little off. The problem is when your finding the probabilites of 2 different subsets of the same set, sometimes the members of the 2 different subsets overlap, and you end up counting them twice. In this case 4,5 and 5,4 are counting both times in P(9) and P(diff1). So the actual probability of Event A OR Event B happening is:
P(A) + P(B) - P(A $\cap$ B)
A $\cap$ B just mean that both are true.

So the P(Both) = $\frac{1}{18}$

So..
$\frac{2}{18}+\frac{5}{18}-\frac{1}{18} = \frac{6}{18} = \frac{1}{3}$