Question

A die is thrown three times. Find the probability of getting a number greater than 4 at least once.

Hint:

Use complement rule: \( P(\text{at least one}) = 1 - P(\text{none}) \). Convert the probability into total favorable and total possible cases.

Answer 1

Total outcomes when a die is thrown 3 times: \[ n(S) = 6 \times 6 \times 6 = 216 \] Let \( A \) be the event: number \( > 4 \) occurs at least once. Numbers > 4 are: 5 and 6 \( \Rightarrow \) favorable outcomes = 2 So, let’s find complement event: no number > 4 That means all outcomes are from {1,2,3,4} So each roll has 4 options: \( 4^3 = 64 \) \[ P(A) = 1 - P(A') = 1 - \frac{64}{216} = \frac{152}{216} = \frac{19}{27} \] Final Answer: \[ \boxed{\frac{19}{27}} \]