A dice is rolled 6 times. The probability of obtaining 2 and 4 exactly...
Understanding the Problem
When rolling a dice 6 times, we want to find the probability of getting the numbers 2 and 4 exactly three times each.
Defining the Scenario
- We have a total of 6 rolls.
- We need to get 2 three times and 4 three times.
- The remaining outcomes can be any of the other four faces (1, 3, 5, 6), which we do not care about.
Calculating the Probability
1. Ways to Choose Positions:
- The number of ways to choose 3 positions for the number 2 from 6 rolls is given by the combination formula:
\[
\binom{6}{3} = 20
\]
- The remaining positions (3 of them) will automatically be for the number 4.
2. Calculating the Outcomes:
- For the other three rolls, each can be any of the faces 1, 3, 5, or 6. Hence, there are:
\[
4^0 = 1 \quad \text{(Since we don't want any of these numbers)}
\]
3. Total Outcomes:
- The total possible outcomes when rolling a dice 6 times is:
\[
6^6 = 46656
\]
4. Probability Calculation:
- Therefore, the probability \( P \) is calculated as:
\[
P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{20 \cdot 1}{46656} = \frac{20}{46656}
\]
5. Final Probability:
- Simplifying \( \frac{20}{46656} \) gives:
\[
\frac{5}{11664}
\]
Conclusion
Thus, the correct answer is option 'C' \( \frac{5}{11664} \).