Question

an integer is chosen between 0 and 100 what is the probability that it is :

(1) divisible by 9

(2) a perfect square number

Answer 1

The list of numbers between 0 and 100 are : 1, 2, 3, 4, 5,...........99
total possible outcomes = 99
numbers which are divisible by 9 are : 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99
no. of favorable outcomes = 11

$$\begin{gathered} \mathrm{probabilty}(\mathrm{no}.\mathrm{is}\mathrm{divisible}\mathrm{by}9)=\frac{\mathrm{number}\mathrm{of}\mathrm{favorable}\mathrm{outcomes}}{\mathrm{no}.\mathrm{of}\mathrm{possible}\mathrm{outcomes}} \\ \Rightarrow \mathrm{probabilty}(\mathrm{no}.\mathrm{is}\mathrm{divisible}\mathrm{by}9)=\frac{11}{99}=\frac{1}{9} \\ \mathrm{Perfect}\mathrm{square}\mathrm{numbers}\mathrm{between}0\mathrm{and}100\mathrm{are}:\{1,4,9,16,25,36,49,64,81\} \\ \mathrm{number}\mathrm{of}\mathrm{perfect}\mathrm{squares}=9 \\ \mathrm{no}.\mathrm{of}\mathrm{favorable}\mathrm{outcomes}=9 \\ \mathrm{probabilty}(\mathrm{the}\mathrm{number}\mathrm{is}\mathrm{perfect}\mathrm{square})=\frac{\mathrm{number}\mathrm{of}\mathrm{favorable}\mathrm{outcomes}}{\mathrm{no}.\mathrm{of}\mathrm{possible}\mathrm{outcomes}} \\ =\frac{9}{99} \\ =\frac{1}{11} \end{gathered}$$