Total number of discs = 90
(i) Let E1 be the event of having a two digit number.
Number of discs bearing two digit number = 90 - 9 = 81
Let E1 be the event of getting a good pen
Therefore P(getting a two digit number) = P(E1) = \(\frac{number\,of\,outcomes\,favorable\,to\,E_1}{number\,of\,all\,possible\,outcomes}\) = \(\frac{81}{90}\) = \(\frac{9}{10}\)
Thus, the probability that the discs bears a two digit number is \(\frac{9}{10}\).
(ii) Let E2 be the event of getting a perfect squaring number.
Disc bearing perfect square numbers are 1,4,9,16,25,36,49,64 and 81.
number of discs bearing a perfect square number = 9
Therefore P(getting a perfect square number) = P(E2) = \(\frac{number\,of\,outcomes\,favorable\,to\,E_2}{number\,of\,all\,possible\,outcomes}\) = \(\frac{9}{90}\) = \(\frac{1}{10}\)
Thus, the probability that the discs bears a perfect square number is \(\frac{1}{10}\)
(iii) Let E3 be the event of getting a number divisible by 5.
Discs bearing numbers divisible by 5 are 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85 and 90.
number of discs bearing a number divisible by 5 = 18
Therefore P(getting a number divisible by 5) = P(E3) = \(\frac{number\,of\,outcomes\,favorable\,to\,E_3}{number\,of\,all\,possible\,outcomes}\) = \(\frac{18}{90}\) = \(\frac{1}{5}\)
Thus, the probability that the discs bears a number divisible by 5 is \(\frac{1}{5}\).