Question

A box contains 90 discs numbered 1 to 90. One disc is drawn at random from the box. What is the probability that is bears

 

(i) a two-digit number

 

(ii) a perfect square

 

(iii) a multiply of 5

 

(iv) a number divisible by 3 and 5.

Answer 1

(i) Total number of possible outcomes = 90 (Since the disks are numbered from 1 to 90).

Number of favourable outcomes of the event E     = Number of two-digit numbers from 1 to 90

  = 90 - 9 = 81 (Since all are two-digit numbers except 1 to 9)

Therefore, P(E) = Number of Favourable Outcomes of the Event ETotal Number of Possible Outcomes

                       = 81/90

                       = 9/10.

 

ii) Total number of possible outcomes = 90.

 

Number of favourable outcomes of the event F

= Number of perfect squares from 1 to 90

 = 9 [Since 1, 4, 9, 16, 25, 36, 49, 64 and 81 are perfect squares).

Therefore, by definition, P(F) = 9/90

                                           = 1/10.

 

 

(iii) Total number of possible outcomes = 90.

Number of favourable outcomes of the event G

= Number of multiples of 5 among numbers from 1 to 90

 = 18 [Since 5 × 1, 5 × 2, 5 × 3, ...., 5 × 18 are multiple of 5).

Therefore, by definition, P(G) = 18/90

                                           = 1/5.

 

(iv) Total number of possible outcomes = 90.

Number of favourable outcomes of the event H

= Number of numbers divisible by 3 and 5 from 1 to 90

= Number of numbers divisible by 15 from 1 to 90

= 6 [Since 15 × 1, 15 × 2, 15 × 3, ...., 15 × 6 are divisible by 15).

Therefore, by definition, P(H) = 6/90

                                           = 1/15.

(i) 9/10

(ii)1/10

(iii)1/5

(iv)1/15