Three different dice are rolled three times. The Probability that they show different numbers only two times is:
Step-by-Step Solution
Key Concept: Recognize that 'different numbers only two times' means exactly 2 out of 3 rolls show all different faces on the three dice. Use binomial approach: C(3,2) × P(exactly 2 rolls with all different faces) × P(exactly 1 roll without all different faces), where P(all different) = 120/216 = 5/9 and P(not all different) = 4/9.
Probability that 3 different dice show different faces is $\frac{P_6}{6^3} = \frac{6!/(6-3)!}{6^3} = \frac{120}{216} = \frac{5}{9}$. Using conditional probability with at least one pair matching: $\binom{3}{2}\left(\frac{5}{9}\right)^2 \times \left(\frac{4}{9}\right) = \frac{100}{243}$.
Correct Answer: 4