A fair die is rolled. The probability of getting a PRIME number is

1. Sample space: $\{1, 2, 3, 4, 5, 6\}$. 2. Primes among them: $2, 3, 5$. (1 is NOT prime by convention.) 3. Event 'prime': $E = \{2, 3, 5\}$, $|E| = 3$. 4. $P(\text{prime}) = 3/6 = 1/2$. 5. Note: this matches the probability of odd (B option of an earlier question). They have a different event SET but same cardinality — coincidence. 6. Option A would correspond to $|E| = 2$ (missing one prime). Option C is for $|E| = 4$. Option D is for excluding only one outcome. _Source: syllabus Mathematics, Ch 14, Example 1 (Prime event on die), p. 3._