A die is thrown twice. What is the probability that:
(i) $5$ will not come up either time?
(ii) $5$ will come up at least once?
Step-by-Step Solution
Key Concept: Total outcomes when throwing a die twice $= 6 \times 6 = 36$. Outcomes where 5 comes up at least once: $(5,1)..(5,6)$ and $(1,5)..(6,5)$ (11 outcomes).
Stepwise Solution:
Total outcomes $= 36$. [0.5 Mark]
Outcomes where 5 comes up at least once $= \{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(1,5),(2,5),(3,5),(4,5),(6,5)\}$ (11 outcomes). [1.0 Mark]
(i) $P(5 \text{ will not come up either time}) = 1 - \dfrac{11}{36} = \dfrac{25}{36}$. [1.0 Mark]
(ii) $P(5 \text{ will come up at least once}) = \dfrac{11}{36}$. [0.5 Mark]
Marking Scheme:
• Sample space 36 and identifying 11 outcomes with 5: 1.5 Marks
• Part (i) $P = 25/36$: 1.0 Mark
• Part (ii) $P = 11/36$: 0.5 Mark
Correct Answer: