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. Fig. 14.4 Fig. 14.5 216
Step-by-Step Solution
Key Concept: Use the definition of probability: $$P(E)=\frac{\text{Number of favourable outcomes}}{\text{Total number of equally likely outcomes}}.$$ Count the favourable numbers for each part and simplify the fraction.
1. Total number of possible outcomes\
The disc is drawn at random from 90 discs, so the total number of equally likely outcomes \(n(S)\) is \(90\).
2. (i) Two‑digit numbers\
- Single‑digit numbers are \(1,2,\dots,9\) (9 numbers).\
- Hence two‑digit numbers are the remaining numbers from \(10\) to \(90\).\
- Count of two‑digit numbers \(=90-9=81\).
- Probability\
$$P(\text{two‑digit})=\frac{81}{90}=\frac{9}{10}.$$
3. (ii) Perfect square numbers\
- List the perfect squares ≤ 90: \(1^2,2^2,3^2,\dots,9^2\) i.e. \(1,4,9,16,25,36,49,64,81\).
- Number of perfect squares \(=9\).
- Probability\
$$P(\text{perfect square})=\frac{9}{90}=\frac{1}{10}.$$
4. (iii) Numbers divisible by 5\
- Multiples of 5 up to 90 are \(5,10,15,\dots,90\).
- Count = \(\frac{90}{5}=18\).
- Probability\
$$P(\text{divisible by 5})=\frac{18}{90}=\frac{1}{5}.$$
5. Final answers\
- (i) \(\displaystyle \frac{9}{10}\)\
- (ii) \(\displaystyle \frac{1}{10}\)\
- (iii) \(\displaystyle \frac{1}{5}\)
Correct Answer: (i) \(\frac{9}{10}\) (ii) \(\frac{1}{10}\) (iii) \(\frac{1}{5}\)