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Home » 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.
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.
CBSE | Class 10 | Exercise 1 | maths | Maths | ML Aggarwal | Probability
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.

Solution:

The possible outcomes are {1,2,3,…90}

Number of possible outcomes = 90

(i) Let E be the event of getting the number on the disc is a two-digit number.

Outcomes favourable to E are {10,11,12,….90}

Number of favourable outcomes = 81

P(E) = 81/90 = 9/10

Hence the probability of getting the number on the disc is a two-digit number is 9/10.

(ii) Let E be the event of getting the number on the disc is a perfect square number.

Outcomes favourable to E are {1,4,9,16,25,36,49,64,81}

Number of favourable outcomes = 9

P(E) = 9/90 = 1/10

Hence the probability of getting the number on the disc is a perfect square number is 1/10.

(iii) Let E be the event of getting the number on the disc is a number divisible by 5.

Outcomes favourable to E are {5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90}

Number of favourable outcomes = 18

P(E) = 18/90 = 2/10 = 1/5

Hence the probability of getting the number on the disc is a number divisible by 5 is 1/5.

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