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A box contains 90 discs which are numbered from 1 to 90. If on

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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 $5$

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A box contains

90

discs which are numbered from

1

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

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Text solutionVerified

i) Number of

2

digit numbers from

1

90 = 81.

P (2

digit number

) = \frac{81}{90} = \frac{9}{10}

ii) Number of perfect squares between

1

and

90 = (1, 4, 9, 16, 25, 36, 49, 64, 81) = 9

P (

Perfect square

) = \frac{9}{90} = \frac{1}{10}

iii) Multiples of

5

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 (

Multiples of

5) = \frac{18}{90} = \frac{1}{5}

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Question Text	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 $5$
Answer Type	Text solution:1
Upvotes	150

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

Text solutionVerified

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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 $5$