A die is thrown once. Find the probability of getting:
(i) a prime number
(ii) a number lying between $2$ and $6$.
Step-by-Step Solution
Key Concept: Total outcomes when throwing a die once $= \{1, 2, 3, 4, 5, 6\}$ (6 outcomes).
Stepwise Solution:
Total outcomes $= 6$.
(i) Prime numbers are $\{2, 3, 5\}$ (3 outcomes).
$P(\text{prime number}) = \dfrac{3}{6} = \dfrac{1}{2}$. [1.0 Mark]
(ii) Numbers between 2 and 6 are $\{3, 4, 5\}$ (3 outcomes).
$P(\text{number between 2 and 6}) = \dfrac{3}{6} = \dfrac{1}{2}$. [1.0 Mark]
Marking Scheme:
• Part (i) Favourable outcomes {2, 3, 5} $\Rightarrow P = 1/2$: 1.0 Mark
• Part (ii) Favourable outcomes {3, 4, 5} $\Rightarrow P = 1/2$: 1.0 Mark
Correct Answer: