Question

A bag contains balls numbered 2 to 91 such that each ball bears a different number. A ball is drawn at random from the bag. Find the probability that:

• [(i)] it bears a 2-digit number
• [(ii)] it bears a multiple of 1

Hint:

All natural numbers are multiples of 1. Always count favourable outcomes carefully within the given range.

Answer 1

Given:
Balls numbered from 2 to 91, each ball with a different number.

Total number of balls:
\[ 91 - 2 + 1 = 90 \]

Part (i): Probability that the ball bears a 2-digit number
Two-digit numbers between 2 and 91 are from 10 to 91.
Number of two-digit numbers = \(91 - 10 + 1 = 82\)
Therefore,
\[ \text{Probability} = \frac{82}{90} = \frac{41}{45} \]

Part (ii): Probability that the ball bears a multiple of 1
Every integer is a multiple of 1.
Therefore, all 90 balls are multiples of 1.
So,
\[ \text{Probability} = \frac{90}{90} = 1 \]

Final Answer:
(i) \(\frac{41}{45}\)
(ii) 1