Final answer:
The probability of throwing at least one six when a fair cubical die is thrown n times is 1 - (5/6)^n. The least value of n for which the probability is 99.5% is approximately 12.
Step-by-step explanation:
To find the probability of throwing at least one six when a fair cubical die is thrown n times, you can use the complement rule: P(throw at least one six) = 1 - P(no six).
Assuming the die is fair, the probability of getting a six on one throw is 1/6. So, the probability of not getting a six on one throw is 5/6. For n throws, the probability of not getting a six on any throw is (5/6)^n. Therefore, the probability of throwing at least one six is 1 - (5/6)^n.
To find the least value of n for which P(throw at least one six in n throws) is 99.5%, we need to solve the equation 1 - (5/6)^n = 0.995.
By taking logarithm on both sides and solving for n, we get n = log(0.005) / log(5/6).
After substituting the values, we find n ≈ 12.