\(\mathbf{1 - P(A^C)^n}\)
How many dice must be rolled to have at least a 95% chance of rolling a six?
Answer
The probability of not rolling a six in a single roll is \(\frac{5}{6}\). Therefore, the probability of not rolling a six in \(n\) rolls is \(\left(\frac{5}{6}\right)^n\). To have at least a 95% chance of rolling at least one six, we set up the inequality:
\[
1 - \left(\frac{5}{6}\right)^n \geq 0.95
\]
\[
\left(\frac{5}{6}\right)^n \leq 0.05
\]
Taking the natural logarithm of both sides:
\[
n \geq \frac{\ln(0.05)}{\ln\left(\frac{5}{6}\right)} \approx 16.43
\]
Since \(n\) must be an integer, at least \(17\) dice must be rolled to achieve a 95% chance of rolling at least one six.
Notes and comments
Comment 1: In such exercises, the key is to express the original probability, $A$, as $1 - P(A^C)^n$, because this allows us to isolate $n$ by using logarithms. This idea will also be used in the next three exercises, as we will see.