From Ross, Intro to Probability Models – Chapter 2.
- Suppose three fair dice are rolled. What is the probability at most one six appears?
Attempted Solution
The total amount of combinations are 216. Find all combinations with at most 1 six:
These will be 6 _ _ -> 25 combinations, _ 6 _ -> 25 combinations. and _ _ 6, which are another set of combinations. These give me 75 combinations. Then, there are the set of combinations where there is not a 6. These can be found by:
$$P(No \space 6s) = \binom{3}{0}(\frac{1}{6})^0(\frac{5}{6})^3 = \frac{125}{216}$$
So then the total probability becomes 200/216.
We can get the same solution by taking the approach from behind: find all cases where there is at least 2 or 3 sixes and subtract from 1:
$$1-\operatorname{nCr}\left(3,2\right)\left(\frac{1}{6}\right)^{2}\left(\frac{5}{6}\right)-\frac{1}{6}^{3}$$
Why is the Ross solution 198/216? Can someone help me out?
Best Answer
Your answer is correct, and the second method is how I would have done it. It appears that there is an error with the book. I wrote a simple python program that checks all possibilities and count when there is $\leq 1$ six (just cause I wanted to confirm I wasn't screwing up my math)
Since there are $200$ cases of at most $1$ six, then the probability is $\frac{200}{216}$