A fair die is rolled eight times. Describe the experiment with a probability space, and find the probability that a 6 is rolled at least once.

Step-by-step solution

1. 1

   Set up the probability space. An outcome records all eight results in order, so

   $\Omega=\{1,\ldots,6\}^{8},\qquad |\Omega|=6^{8}=1\,679\,616,$

   with the uniform measure, since the die is fair and the rolls are independent.

2. 2

   Turn 'at least once' into its complement. Counting the ways to get one, two, ... up to eight sixes would mean summing eight binomial terms. The complement is a single event:

   $P(\text{at least one }6)=1-P(\text{no }6\text{ at all}).$

   'At least one' is almost always a signal to use the complement.

3. 3

   Compute the probability of no six. Each individual roll avoids a $6$ with probability $\tfrac56$, and the rolls are independent, so all eight avoid it with probability

   $\left(\frac{5}{6}\right)^{8}=\frac{390\,625}{1\,679\,616}=0.232568.$

4. 4

   Subtract from 1.

   $P(\text{at least one }6)=1-\frac{390\,625}{1\,679\,616}=\frac{1\,288\,991}{1\,679\,616}\approx 0.7674.$

5. 5

   Check the answer is plausible. The expected number of sixes is $8\times\tfrac16=1.33$, so seeing at least one should be likely but not near-certain — and $76.7\%$ fits that. Note it is not $8\times\tfrac16=133\%$; adding probabilities of overlapping events is the classic error this complement approach avoids.