A fair die is rolled four times. Describe the experiment with a probability space, and find the probability that a 1 is rolled at most three times.

Step-by-step solution

1. 1

   Build the probability space. An outcome is the ordered list of the four results, so the sample space is

   $\Omega=\{1,2,3,4,5,6\}^{4},\qquad |\Omega|=6^{4}=1296.$

   The die is fair and the rolls are independent, so the measure is uniform: every one of the $1296$ outcomes has probability $\tfrac{1}{1296}$.

2. 2

   Rewrite the event using its complement. Let $X$ be the number of $1$s among the four rolls. 'At most three' means $X\in\{0,1,2,3\}$ — four separate cases. Its complement is the single case $X=4$:

   $P(X\le 3)=1-P(X=4).$

   Computing one term instead of four is the whole reason to use the complement here.

3. 3

   Count the outcomes in the complement. $X=4$ means every roll shows a $1$, which is exactly one outcome, $(1,1,1,1)$:

   $P(X=4)=\left(\frac{1}{6}\right)^{4}=\frac{1}{1296}.$

4. 4

   Subtract.

   $P(X\le 3)=1-\frac{1}{1296}=\frac{1295}{1296}\approx 0.99923.$

5. 5

   Sanity-check the size of the answer. Getting four $1$s in a row is very unlikely, so 'at most three' should be almost certain — and $0.99923$ is. As a cross-check, the binomial model $X\sim B(4,\tfrac16)$ gives $P(X=4)=\binom{4}{4}\left(\tfrac16\right)^4\left(\tfrac56\right)^0=\tfrac{1}{1296}$, the same value.