A fair six-sided die is rolled 6 times. What is the probability of 5 and 6 appearing at least once?

Question

My approach was to calculate the probability of not throwing a 5 or a 6, and to time that by the number of throws. 4/6^6 = 0.09 and thus the opposing probability would be 91%. That, unfortunately, tells me only the odds of rolling a 5 OR a 6 in six rolls, not the wanted probability of 5 AND 6. How should I change my approach?

Answer 1

Let the event $A$ be "there are no '5' in 6 throws" and the event $B$ be "there are no '6' in 6 throws". Then the probability in question is:

$$1-P(A)-P(B)+P(A\cap B)=1-2\left(\frac56\right)^6+\left(\frac46\right)^6 \approx0.418.$$

Answer 2

The Principle of Inclusion-Exclusion says the probability that, after $n$ rolls, a $5$ and a $6$ have been rolled is $$ P(n)=1-\left(\vphantom{\frac56^n}\right.\overbrace{\ \ \left(\frac56\right)^n\ \ }^\text{no $5$s}+\overbrace{\ \ \left(\frac56\right)^n\ \ }^\text{no $6$s}-\overbrace{\ \ \left(\frac46\right)^n\ \ }^\text{no $5$s or $6$s}\left.\vphantom{\frac56^n}\right) $$ $$ \begin{array}{c|c|r} n&P(n)&\%\phantom{00}\\\hline 0&0&0.00\\ 1&0&0.00\\ 2&\frac{1}{18}&5.56\\ 3&\frac{5}{36}&13.89\\ 4&\frac{151}{648}&23.30\\ 5&\frac{425}{1296}&32.79\\ \color{#C00}{6}&\color{#C00}{\frac{9751}{23328}}&\color{#C00}{41.80}\\ 7&\frac{23345}{46656}&50.04 \end{array} $$