In a throw of three dice, what is the probability of getting one prime number, one composite number and one number which is <strong>NEITHER</strong> prime <strong>NOR</strong> composite?

Explanation

A die has faces \{1, 2, 3, 4, 5, 6\}. The primes \{2, 3, 5\} provide 3 outcomes. The composites \{4, 6\} provide 2 outcomes. The number that is neither \{1\} provides 1 outcome. The number of ways to arrange these three distinct types across 3 dice is 3! = 6. The total favorable outcomes is 3 \times 2 \times 1 \times 6 = 36. The probability is \frac{36}{6^3} = \frac{1}{6}.

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