Suppose three fair dice are rolled. the probability at most one six appears

diceprobabilityprobability theory

From Ross, Intro to Probability Models – Chapter 2.

  1. 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)

count = 0
for i in range(1,7):
    for j in range(1,7):
        for k in range(1,7):
            if ((i==6) + (j==6) + (k==6)) <= 1:
                count += 1
print(count)
# Output is 200

Since there are $200$ cases of at most $1$ six, then the probability is $\frac{200}{216}$

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