Question

If a die is thrown three times, then find the probability of getting at least one odd number.

Hint:

Use the complement: 1 - P(all three throws are even).

Answer 1

Step 1: List why direct counting of \"at least one\" is harder than its complement:
\"At least one odd\" includes exactly-one, exactly-two, and all-three-odd cases — three cases to add up directly.

Step 2: The complement (\"zero odd\", i.e. all even) is a single simple case:
Each throw independently must land even, probability $1/2$ each time, so all three: $(1/2)^3=1/8$.

Step 3: Subtract from 1:
$P(\text{at least one odd})=1-1/8=7/8$.

Final Answer:
\[ \boxed{7/8} \]