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# 18. 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.

Pawan Prajapati
3 months ago
The total numbers of discs = 50 P(E) = (Number of favourable outcomes/ Total number of outcomes) (i) Total number of discs having two digit numbers = 81 (Since 1 to 9 are single digit numbers and so, total 2 digit numbers are 90-9 = 81) P (bearing a two-digit number) = 81/90 = 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) = 9/90 = 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) = 18/90 = ⅕ = 0.2
Armaan Arora
100 Points
2 months ago
The total numbers of discs = 50 P(E) = (Number of favourable outcomes/ Total number of outcomes) (i) Total number of discs having two digit numbers = 81 (Since 1 to 9 are single digit numbers and so, total 2 digit numbers are 90-9 = 81) P (bearing a two-digit number) = 81/90 = 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) = 9/90 = 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) = 18/90 = ⅕ = 0.2

by armaan