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.
Total number of discs = 90. Two-digit numbers from 1 to 90 are 10 to 90. Number of two-digit numbers = 90 - 9 = 81. P(two-digit number) = 81/90 = 9/10.
Perfect squares from 1 to 90 are: 1, 4, 9, 16, 25, 36, 49, 64, 81. Number of perfect squares = 9. P(perfect square) = 9/90 = 1/10.
Numbers divisible by 5 from 1 to 90 are: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90. Number of multiples of 5 = 18. P(number divisible by 5) = 18/90 = 1/5.
Explanation
This is Question 18 from Exercise 15.1 of NCERT Class 10 Mathematics textbook. The question tests the fundamental probability formula P(E) = Number of favorable outcomes / Total number of possible outcomes. For part (i), two-digit numbers exclude single-digit numbers (1-9). For part (ii), perfect squares are identified by squaring integers 1 through 9. For part (iii), multiples of 5 are found by dividing 90 by 5.