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
(iii) A number divisible by $5$
Step-by-Step Solution
Key Concept: (i) Two-digit: $10$ to $90 = 81$ numbers $\Rightarrow P = 81/90 = 9/10$.<br>(ii) Perfect squares: $1, 4, 9, 16, 25, 36, 49, 64, 81 = 9$ numbers $\Rightarrow P = 9/90 = 1/10$.<br>(iii) Divisible by 5: $5, 10, 15, \dots, 90 = 18$ numbers $\Rightarrow P = 18/90 = 1/5$.
(i) $P(\text{Two-digit}) = 81/90 = 9/10$. [1.0 Mark]
(ii) $P(\text{Perfect Square}) = 9/90 = 1/10$. [1.0 Mark]
(iii) $P(\text{Divisible by 5}) = 18/90 = 1/5$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Part (i) $P = 9/10$: 1.0 Mark
Part (ii) $P = 1/10$: 1.0 Mark
Part (iii) $P = 1/5$: 1.0 Mark
Correct Answer: