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

Total number of discs = 90

(i) Total number of two-digit numbers between 1 and 90 = 81

P (getting a two-digit number) = 81/90 =9/10

(ii) Perfect squares between 1 and 90 are 1, 4, 9, 16, 25, 36, 49, 64, and 81. Therefore, total number of perfect squares between 1 and 90 is 9.

P (getting a perfect square) = 9/90 = 1/10

(iii) Numbers that are between 1 and 90 and divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, and 90. Therefore, total numbers divisible by 5 = 18

Probability of getting a number divisible by 5 =18/90 = 1/5

(i) 10,11,12 … 90 are two digit numbers. There are 81 numbers. So, Probability of getting a two-digit number 

\(\frac{81}{90}=\frac{9}{10}\)

(ii) 1, 4, 9,16,25,36,49,64,81 are perfect squares. So, Probability of getting a perfect square number. 

\(\frac{9}{90}=\frac{1}{10}\)

(iii) 5, 10,15 ... 90 are divisible by 5. There are 18 outcomes. So,Probability of getting a number divisible by 5. 

= \(\frac{18}{90}=\frac{1}{5}\)

(i) Number of two digit numbers = 90 – (1, 2, 3, 4, 5, 6, 7, 8, 9) = 81

P(a two-digit number) = 81/90

(ii) Number of perfect squares = 1, 4, 9, 16, 25, 36, 49, 64, 81 = 9

P(a perfect square number) = 9/90

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

P(a number divisible by 5) = 18/90