A box contains $90$ discs which are numbered from $1$ to $90$. If one disc is drawn at random from the box, find the probability that it bears:
(i) a two-digit number
(ii) a perfect square number.
Step-by-Step Solution
Key Concept: Total outcomes $= 90$. Two-digit numbers: 10 to 90 (81 numbers). Perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81 (9 numbers).
Stepwise Solution:
(i) Two-digit numbers are $10$ to $90$ ($81$ numbers).
$P(\text{two-digit number}) = \dfrac{81}{90} = \dfrac{9}{10} = 0.9$. [1.0 Mark]
(ii) Perfect squares are $\{1, 4, 9, 16, 25, 36, 49, 64, 81\}$ ($9$ numbers).
$P(\text{perfect square}) = \dfrac{9}{90} = \dfrac{1}{10} = 0.1$. [1.0 Mark]
Marking Scheme:
• Part (i) 81 two-digit numbers $\Rightarrow P = 9/10$: 1.0 Mark
• Part (ii) 9 perfect square numbers $\Rightarrow P = 1/10$: 1.0 Mark
Correct Answer: