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# 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.

The total numbers of discs = 50

$$P(E) \ = \ \frac{Number \ of \ favourable \ outcomes}{Total \ number \ of \ outcomes}$$

(i) Total number of discs having two digit numbers = 81

($$\because$$ 1 to 9 are single digit numbers and so, total 2 digit numbers are 90 - 9 = 81)

P (bearing a two-digit number) $$= \ \frac{81}{90} \ = \ \frac{9}{10} \ = \ 0.9$$

(ii) Total number of perfect square numbers = 9 (1, 4, 9, 16, 25, 36, 49, 64 and 81)

P (getting a perfect square number) $$=\ \frac{9}{90} \ = \ \frac{1}{10} \ = \ 0.1$$

(iii) Total numbers which are divisible by 5 = 18 (5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90)

P (getting a number divisible by 5) $$= \ \frac{18}{90} \ = \ \frac{1}{5} \ = \ 0.2$$