Question

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

Hint:

In probability problems involving "at least one", it is easier to use the complement rule: $P(\text{at least one}) = 1 - P(\text{none})$.

Answer 1

Step 1: Define the experiment.
A die is thrown three times. Each throw has $6$ possible outcomes. Total number of outcomes: \[ n(S) = 6^3 = 216 \]

Step 2: Use complementary probability.
We want the probability of at least one odd number. \[ P(\text{at least one odd}) = 1 - P(\text{no odd}) \]

Step 3: Probability of no odd number.
Odd numbers on a die: $\{1,3,5\}$ (3 outcomes). Even numbers on a die: $\{2,4,6\}$ (3 outcomes). Probability of even in one throw = $\dfrac{3}{6} = \dfrac{1}{2}$. For three throws, probability of all even = \[ \left(\frac{1}{2}\right)^3 = \frac{1}{8} \]

Step 4: Final probability.
\[ P(\text{at least one odd}) = 1 - \frac{1}{8} = \frac{7}{8} \]

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